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Section 6.1 Using Definite Integrals to Find Area and Length

Early on in our work with the definite integral, we learned that for an object moving along an axis, the area under a non-negative velocity function \(v\) between \(a\) and \(b\) tells us the distance the object traveled on that time interval, and that area is given precisely by the definite integral \(\int_a^b v(t) \, dt\text{.}\) In general, for any nonnegative function \(f\) on an interval \([a,b]\text{,}\) \(\int_a^b f(x) \, dx\) measures the area bounded by the curve and the \(x\)-axis between \(x = a\) and \(x = b\text{.}\)

Next, we will explore how definite integrals can be used to represent other physically important properties. In Preview Activity 6.1.1, we investigate how a single definite integral may be used to represent the area between two curves.

Preview Activity 6.1.1.

Consider the functions given by \(f(x) = 5-(x-1)^2\) and \(g(x) = 4-x\text{.}\)

  1. Use algebra to find the points where the graphs of \(f\) and \(g\) intersect.

  2. Sketch an accurate graph of \(f\) and \(g\) on the axes provided, labeling the curves by name and the intersection points with ordered pairs.

  3. Find and evaluate exactly an integral expression that represents the area between \(y = f(x)\) and the \(x\)-axis on the interval between the intersection points of \(f\) and \(g\text{.}\)

  4. Find and evaluate exactly an integral expression that represents the area between \(y = g(x)\) and the \(x\)-axis on the interval between the intersection points of \(f\) and \(g\text{.}\)

  5. What is the exact area between \(f\) and \(g\) between their intersection points? Why?

Figure 6.1.1. Axes for plotting \(f\) and \(g\) in Preview Activity 6.1.1

Subsection 6.1.1 The Area Between Two Curves

In Preview Activity 6.1.1, we saw a natural way to think about the area between two curves: it is the area beneath the upper curve minus the area below the lower curve.

Find the area bounded between the graphs of \(f(x) = (x-1)^2 + 1\) and \(g(x) = x+2\text{.}\)

Figure 6.1.3. The areas bounded by the functions \(f(x) = (x-1)^2 + 1\) and \(g(x) = x+2\) on the interval \([0,3]\text{.}\)
Solution

In Figure 6.1.3, we see that the graphs intersect at \((0,2)\) and \((3,5)\text{.}\) We can find these intersection points algebraically by solving the system of equations given by \(y = x+2\) and \(y = (x-1)^2 + 1\text{:}\) substituting \(x+2\) for \(y\) in the second equation yields \(x+2 = (x-1)^2 + 1\text{,}\) so \(x+2 = x^2 - 2x + 1 + 1\text{,}\) and thus

\begin{equation*} x^2 - 3x = x(x-3) = 0\text{,} \end{equation*}

from which it follows that \(x = 0\) or \(x = 3\text{.}\) Using \(y = x+2\text{,}\) we find the corresponding \(y\)-values of the intersection points.

On the interval \([0,3]\text{,}\) the area beneath \(g\) is

\begin{equation*} \int_0^3 (x+2) \, dx = \frac{21}{2}\text{,} \end{equation*}

while the area under \(f\) on the same interval is

\begin{equation*} \int_0^3 [(x-1)^2 + 1] \, dx = 6\text{.} \end{equation*}

Thus, the area between the curves is

\begin{equation} A = \int_0^3 (x+2) \, dx - \int_0^3 [(x-1)^2 + 1] \, dx = \frac{21}{2} - 6 = \frac{9}{2}\text{.}\label{wfW}\tag{6.1.1} \end{equation}

We can also think of the area this way: if we slice up the region between two curves into thin vertical rectangles (in the same spirit as we originally sliced the region between a single curve and the \(x\)-axis in Section 4.2), we see (as shown in Figure 6.1.4) that the height of a typical rectangle is given by the difference between the two functions, \(g(x) - f(x)\text{,}\) and its width is \(\Delta x\text{.}\) Thus the area of the rectangle is

\begin{equation*} A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{.} \end{equation*}
Figure 6.1.4. The area bounded by the functions \(f(x) = (x-1)^2 + 1\) and \(g(x) = x+2\) on the interval \([0,3]\text{.}\)

The area between the two curves on \([0,3]\) is thus approximated by the Riemann sum

\begin{equation*} A \approx \sum_{i=1}^{n} (g(x_i) - f(x_i)) \Delta x\text{,} \end{equation*}

and as we let \(n \to \infty\text{,}\) it follows that the area is given by the single definite integral

\begin{equation} A = \int_0^3 (g(x) - f(x)) \, dx\text{.}\label{oBx}\tag{6.1.2} \end{equation}

In many applications of the definite integral, we will find it helpful to think of a “representative slice” and use the definite integral to add these slices. Here, the integral sums the areas of thin rectangles.

Finally, it doesn't matter whether we think of the area between two curves as the difference between the area bounded by the individual curves (as in (6.1.1)) or as the limit of a Riemann sum of the areas of thin rectangles between the curves (as in (6.1.2)). These two results are the same, since the difference of two integrals is the integral of the difference:

\begin{equation*} \int_0^3 g(x) \, dx - \int_0^3 f(x) \, dx = \int_0^3 (g(x) - f(x)) \, dx\text{.} \end{equation*}

Our work so far in this section illustrates the following general principle.

If two curves \(y = g(x)\) and \(y = f(x)\) intersect at \((a,g(a))\) and \((b,g(b))\text{,}\) and for all \(x\) such that \(a \le x \le b\text{,}\) \(g(x) \ge f(x)\text{,}\) then the area between the curves is \(A = \int_a^b (g(x) - f(x)) \, dx\text{.}\)

Activity 6.1.2.

In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area.

  1. The finite region bounded by \(y = \sqrt{x}\) and \(y = \frac{1}{4}x\text{.}\)

  2. The finite region bounded by \(y = 12-2x^2\) and \(y = x^2 - 8\text{.}\)

  3. The area bounded by the \(y\)-axis, \(f(x) = \cos(x)\text{,}\) and \(g(x) = \sin(x)\text{,}\) where we consider the region formed by the first positive value of \(x\) for which \(f\) and \(g\) intersect.

  4. The finite regions between the curves \(y = x^3-x\) and \(y = x^2\text{.}\)

Hint

  1. Find where the two curves intersect.

  2. Plot both functions and think about how they compare to \(y=x^2\text{.}\)

  3. Remember that \(\sin(\pi/4)=\cos(\pi/4)\text{.}\)

  4. Find three points where the curves intersect and thus two finite regions between two pairs of those points.

Answer

  1. \(A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx = \frac{32}{3}\text{.}\)

  2. \(A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-x^2)-(x^2-8)) \, dx \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}\)

  3. \(A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.}\)

  4. The left-hand region has area

    \begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514\text{.} \end{equation*}

    The right-hand region has area

    \begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*}
Solution

  1. By solving \(\sqrt{x}=\frac{1}{4}x\text{,}\) we see that the two curves intersect when \(x=0\) and \(x=16\text{,}\) so at the points \((0,0)\) and \((16,4)\text{.}\) We can plot the region as shown in the figure below.

    Slicing the region with thin vertical rectangles, we see that a typical rectangle's height is given by \(\sqrt{x}-\frac{1}{4}x\text{,}\) and thus the area of such a slice is \(A_{\text{rect} } = (\sqrt{x} - \frac{1}{4}x) \Delta x\text{.}\) It follows that the area of the region bounded by the two curves is

    \begin{equation*} A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx\text{.} \end{equation*}

    Evaluating the integral, we find \(A = \frac{32}{3}\text{.}\)

  2. Solving \(12-2x^2=x^2-8\text{,}\) we find that the two curves intersect where \(3x^2=2-\text{,}\) so at \(x=\pm \sqrt{20/3}\text{.}\) We can plot the region as shown in the figure below.

    Slicing the region with thin vertical rectangles, a typical rectangle's height is given by \((12-x^2)-(x^2-8)\text{,}\) and thus the area of such a slice is \(A_{\text{rect} } = ((12-x^2)-(x^2-8)) \Delta x\text{.}\) Hence, the area of the region bounded by the two curves is

    \begin{equation*} A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-x^2)-(x^2-8)) \, dx\text{.} \end{equation*}

    Evaluating the integral, we find \(A = \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}\)

  3. The first positive value at which the sine and cosine functions intersect is \(x=\frac{\pi}{4}\text{.}\) Plotting the functions in a surrounding region, we see the following figure.

    Slicing the region with vertical rectangles, we find the area of the shaded region is given by

    \begin{equation*} A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.} \end{equation*}

  4. The curves \(y = x^3-x\) and \(y = x^2\) intersect at three points, forming two regions. Setting \(x^3-x=x^2\) and solving, we find that

    \begin{equation*} x^3-x^2-x=x(x^2-x-1)=0\text{,} \end{equation*}

    which implies that \(x=0\) or \(x = \frac{1}{2}(1 \pm \sqrt{5})\text{.}\) Plotting the curves, we see that for the left-hand region, the relative interval is \([\frac{1-\sqrt{5}}{2},0]\) and \(g(x)=x^3-x \gt x^2 = f(x)\text{.}\)

    Thus, the left-hand region has area

    \begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514\text{.} \end{equation*}

    For the right-hand region, the interval is \([0, \frac{1+\sqrt{5}}{2}]\) and \(f(x)=x^2 \gt x^3-x = g(x)\text{,}\) so that its area is

    \begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*}

Subsection 6.1.2 Finding Area with Horizontal Slices

At times, the shape of a region may dictate that we use horizontal rectangular slices, instead of vertical ones.

Find the area of the region bounded by the parabola \(x = y^2 - 1\) and the line \(y = x-1\text{,}\) shown at left in Figure 6.1.6.

Figure 6.1.6. The area bounded by the functions \(x = y^2-1\) and \(y = x-1\) (at left), with the region sliced vertically (center) and horizontally (at right).
Solution

By solving the second equation for \(x\) and writing \(x = y + 1\text{,}\) we find that \(y+1 = y^2 - 1\text{.}\) Hence the curves intersect where \(y^2 - y - 2 = 0\text{.}\) Thus, we find \(y = -1\) or \(y = 2\text{,}\) so the intersection points of the two curves are \((0,-1)\) and \((3,2)\text{.}\)

If we attempt to use vertical rectangles to slice up the area (as in the center graph of Figure 6.1.6), we see that from \(x = -1\) to \(x = 0\) the curves that bound the top and bottom of the rectangle are one and the same. This suggests, as shown in the rightmost graph in the figure, that we try using horizontal rectangles.

Note that the width of a horizontal rectangle depends on \(y\text{.}\) Between \(y = -1\) and \(y = 2\text{,}\) the right end of a representative rectangle is determined by the line \(x = y+1\text{,}\) and the left end is determined by the parabola, \(x = y^2-1\text{.}\) The thickness of the rectangle is \(\Delta y\text{.}\)

Therefore, the area of the rectangle is

\begin{equation*} A_{\text{rect} } = [(y+1) - (y^2-1)] \Delta y\text{,} \end{equation*}

and the area between the two curves on the \(y\)-interval \([-1,2]\) is approximated by the Riemann sum

\begin{equation*} A \approx \sum_{i=1}^{n} [(y_i+1)-(y_i^2-1)] \Delta y\text{.} \end{equation*}

Taking the limit of the Riemann sum, it follows that the area of the region is

\begin{equation} A = \int_{y=-1}^{y=2} [(y+1) - (y^2-1)] \, dy\text{.}\label{Neh}\tag{6.1.3} \end{equation}

We emphasize that we are integrating with respect to \(y\text{;}\) this is because we chose to use horizontal rectangles whose widths depend on \(y\) and whose thickness is denoted \(\Delta y\text{.}\) It is a straightforward exercise to evaluate the integral in Equation (6.1.3) and find that \(A = \frac{9}{2}\text{.}\)

Just as with the use of vertical rectangles of thickness \(\Delta x\text{,}\) we have a general principle for finding the area between two curves, which we state as follows.

If two curves \(x = g(y)\) and \(x = f(y)\) intersect at \((g(c),c)\) and \((g(d),d)\text{,}\) and for all \(y\) such that \(c \le y \le d\text{,}\) \(g(y) \ge f(y)\text{,}\) then the area between the curves is

\begin{equation*} A = \int_{y=c}^{y=d} (g(y) - f(y)) \, dy\text{.} \end{equation*}
Activity 6.1.3.

In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area. Note well: At the step where you draw a representative slice, you need to make a choice about whether to slice vertically or horizontally.

  1. The finite region bounded by \(x=y^2\) and \(x=6-2y^2\text{.}\)

  2. The finite region bounded by \(x=1-y^2\) and \(x = 2-2y^2\text{.}\)

  3. The area bounded by the \(x\)-axis, \(y=x^2\text{,}\) and \(y=2-x\text{.}\)

  4. The finite regions between the curves \(x=y^2-2y\) and \(y=x\text{.}\)

Hint

  1. Find where the curves intersect and plot them.

  2. Observe that horizontal slices are appropriate, so we will integrate with respect to y.

  3. Note that the region is approximately triangular with the \(x\)-axis as the base, \(y=x^2\) as the left "side", and \(y = 2-x\) as the right side.

  5. Plot the region and observe that horizontal slices are appropriate because of the shape of the region.

Answer

  1. \(A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.}\)

  2. \(A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \frac{4}{3}\text{.}\)

  3. \(A=\int_{y=0}^{y=4-2\sqrt{3}} \left(\frac{2-y}{2} - \sqrt{y} \right) \, dy = \frac{11}{3}-2\sqrt{3} \approx 0.2026 \)

  4. \(A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \frac{9}{2}\text{.}\)

Solution

  1. The two curves intersect when \(y^2=6-2y^2\text{,}\) thus when \(3y^2=6\text{,}\) so at \(y=\pm \sqrt{2}\text{.}\) Plotting the region, we see that for \(y=-\sqrt{2}\) to \(y=\sqrt{2}\text{,}\) \(g(y) = 6-2y^2 \gt y^2 = f(y)\text{,}\) and thus horizontal slices are appropriate for setting up the integral that represents the region's area. The area is thus given by

    \begin{equation*} A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.} \end{equation*}

  2. The two curves intersect when \(1-y^2=2-2y^2\text{,}\) and therefore \(3y^2=3\) so \(y=\pm 1\text{.}\) Graphing the region, on \(-1 \lt y \lt 1\text{,}\) we see that \(g(y) = 2-2y^2 \ge 1-y^2 = f(y)\text{.}\) Using horizontal slices, it follows that the region's area is given by

    \begin{equation*} A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = int_{y=-1}^{y=1} (1-y^2) \, dy = \frac{4}{3}\text{.} \end{equation*}

  3. Standard algebra shows that the two curves meet at \((-1+\sqrt{3},4-2\sqrt{3})\text{.}\) A careful plot of the region shows that the two curves, together with the \(x\)-axis, form a roughly triangular region with the \(x\)-axis as the base, \(y=x^2\) as the left "side", and \(y = 2-2x\) as the right side. Using horizontal slices, and writing each respective equation with \(x\) as a function of \(y\) (so \(x=\sqrt{y}\) and \(x=\frac{2-y}{2}\)), we find that the area of this region is

    \begin{equation*} A=\int_{y=0}^{y=4-2\sqrt{3}} \left(\frac{2-y}{2} - \sqrt{y} \right) \, dy = \frac{11}{3}-2\sqrt{3} \approx 0.2026 \end{equation*}

  4. The two curves intersect at \((0,0)\) and \((3,3)\text{.}\) For \(0 \le y \le 3\text{,}\) we observe that \(g(y)=y \ge y^2-2y = f(y)\text{,}\) so using horizontal slices and the corresponding definite integral, we find that the region's area is

    \begin{equation*} A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \int_{0}^{3} (3y - y^2) \, dy = \frac{9}{2}\text{.} \end{equation*}

Subsection 6.1.3 Finding the length of a curve

We can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Specifically, we subdivide the curve into small approximating line segments, as shown at left in Figure 6.1.7.

Figure 6.1.7. At left, a continuous function \(y = f(x)\) whose length we seek on the interval \(a = x_0\) to \(b = x_3\text{.}\) At right, a close up view of a portion of the curve.

We estimate the length \(L_{\text{slice} }\) of each portion of the curve on a small interval of length \(\Delta x\text{.}\) We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure 6.1.7. The length, \(h\text{,}\) of the hypotenuse approximates the length, \(L_{\text{slice} }\text{,}\) of the curve between the two selected points. Thus,

\begin{equation*} L_{\text{slice} } \approx h = \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\text{.} \end{equation*}

Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of \((\Delta x)^2\text{,}\) we find

\begin{align*} L_{\text{slice}} &\approx \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\\ &= \sqrt{ (\Delta x)^2\left(1 + \frac{(\Delta y)^2}{(\Delta x)^2} \right)}\\ &= \sqrt{1 + \frac{(\Delta y)^2}{(\Delta x)^2} } \cdot \Delta x\text{.} \end{align*}

Then, as \(n \to \infty\) and \(\Delta x \to 0\text{,}\) we have that \(\frac{\Delta y}{\Delta x} \to \frac{dy}{dx} = f'(x)\text{.}\) Thus, we can say that

\begin{equation*} L_{\text{slice} } \approx \sqrt{1 + f'(x)^2} \Delta x\text{.} \end{equation*}

Taking a Riemann sum of all of these slices and letting \(n \to \infty\text{,}\) we arrive at the following fact.

Given a differentiable function \(f\) on an interval \([a,b]\text{,}\) the total arc length, \(L\text{,}\) along the curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by

\begin{equation*} L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.} \end{equation*}
Activity 6.1.4.

Each of the following questions somehow involves the arc length along a curve.

  1. Use the definition and appropriate computational technology to determine the arc length along \(y = x^2\) from \(x = -1\) to \(x = 1\text{.}\)

  2. Find the arc length of \(y = \sqrt{4-x^2}\) on the interval \(-2 \le x \le 2\text{.}\) Find this value in two different ways: (a) by using a definite integral, and (b) by using a familiar property of the curve.

  3. Determine the arc length of \(y = xe^{3x}\) on the interval \([0,1]\text{.}\)

  4. Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why?

  5. A moving particle is traveling along the curve given by \(y = f(x) = 0.1x^2 + 1\text{,}\) and does so at a constant rate of 7 cm/sec, where both \(x\) and \(y\) are measured in cm (that is, the curve \(y = f(x)\) is the path along which the object actually travels; the curve is not a “position function”). Find the position of the particle when \(t = 4\) sec, assuming that when \(t = 0\text{,}\) the particle's location is \((0,f(0))\text{.}\)

Hint

  1. Recall that \(L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.}\)

  2. Remember that \(x^2 + y^2 = 4\) generates a circle centered at \((0,0)\) of radius 4.

  3. Apply the arc length formula. Use technology to evaluate the integral.

  4. Most expressions involving square roots are difficult to antidifferentiate.

  5. Here you can determine the arc length and then experiment to find the upper limit of integration, which will help you determine position.

Answer

  1. \(L \approx 2.95789\text{.}\)

  2. \(L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}\)

  3. \(L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}\)

  4. We will usually have to estimate the value of \(\int_a^b \sqrt{1+f'(x)^2} \, dx\) using computational technology.

  5. Approximately \((9.011,f(9.011)) = (9.011, 9.1198)\text{.}\)

Solution

  1. Using the formula for arc length, we know

    \begin{equation*} L = \int_{-1}^{1} \sqrt{1+(2x)^2} \,dx\text{.} \end{equation*}

    Evaluating the integral, we find \(L \approx 2.95789\text{.}\)

  2. The curve is a semi-circle of radius 2, so its length must be \(L = \frac{1}{2} \cdot 2 \pi r = 2\pi\text{.}\) This is confirmed by the arc length formula, since \(f(x) = \sqrt{4-x^2}\text{,}\) so \(f'(x) = \frac{1}{2}(4-x^2)^{-1/2}(-2x)\)

    \begin{equation*} L = \int_{-2}^{2} \sqrt{1 + \left(-x(4-x^2)^{-1/2} \right)^2} \, dx = \int_{-2}^{2} \sqrt{1 + x^2(4-x^2)^{-1}} \, dx\text{.} \end{equation*}

    It follows 1  that

    \begin{equation*} L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.} \end{equation*}

  3. For \(y = xe^{3x}\text{,}\) we observe that \(y' = 3xe^{3x} + e^{3x} = e^{3x}(3x+1)\text{,}\) and that

    \begin{equation*} (y')^2 = e^{6x}(9x^2 + 6x + 1)\text{.} \end{equation*}

    Thus, the arc length of the curve on \([0,1]\) is

    \begin{equation*} L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.} \end{equation*}

  4. Because \(\sqrt{1+f'(x)^2}\) will rarely simplify nicely and rarely have an elementary antiderivative, we will usually have to estimate the value of \(\int_a^b \sqrt{1+f'(x)^2} \, dx\) using computational technology.

  5. After 4 seconds traveling at 7 cm/sec, the particle has moved 28 cm. So, we have to find the value of \(b\) for which the arc length \(L\) along \([0,b]\) is 28. Thus, we know that

    \begin{equation*} 28 = L = \int_0^b \sqrt{1 + (0.2x)^2} \, dx\text{.} \end{equation*}

    Rewriting the integral, we see

    \begin{equation*} 28 = \int_0^b \sqrt{1+0.4x^2} \, dx\text{.} \end{equation*}

    Experimenting with different values of \(b\) in a computational engine like Wolfram|Alpha, we find that for \(b \approx 9.011\text{,}\) \(\int_0^{9.011} \sqrt{1+0.4x^2} \, dx \approx 27.9992\text{,}\) and thus the position of the particle when \(t=4\) is approximately \((9.011,f(9.011)) = (9.011, 9.1198)\text{.}\)

This integral is actually "improper" because the integrand is undefined at the endpoints, \(x = \pm 2\text{.}\) We will learn how to evaluate such integrals in Section 6.5.

Subsection 6.1.4 Summary

  • To find the area between two curves, we think about slicing the region into thin rectangles. If, for instance, the area of a typical rectangle on the interval \(x = a\) to \(x = b\) is given by \(A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{,}\) then the exact area of the region is given by the definite integral

    \begin{equation*} A = \int_a^b (g(x)-f(x))\, dx\text{.} \end{equation*}

  • The shape of the region usually dictates whether we should use vertical rectangles of thickness \(\Delta x\) or horizontal rectangles of thickness \(\Delta y\text{.}\) We want the height of the rectangle given by the difference between two curves: if those curves are best thought of as functions of \(y\text{,}\) we use horizontal rectangles, whereas if those curves are best viewed as functions of \(x\text{,}\) we use vertical rectangles.

  • The arc length, \(L\text{,}\) along the curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by

    \begin{equation*} L = \int_a^b \sqrt{1 + f'(x)^2} \, dx\text{.} \end{equation*}

Exercises 6.1.5 Exercises

1. Area between two power functions.
2. Area between two trigonometric functions.
3. Area between two curves.
4. Arc length of a curve.
5.

Find the exact area of each described region.

  1. The finite region between the curves \(x = y(y-2)\) and \(x=-(y-1)(y-3)\text{.}\)

  2. The region between the sine and cosine functions on the interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\text{.}\)

  3. The finite region between \(x = y^2 - y - 2\) and \(y = 2x-1\text{.}\)

  4. The finite region between \(y = mx\) and \(y = x^2-1\text{,}\) where \(m\) is a positive constant.

Answer

  1. \(A = \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy = \sqrt{3} \text{.}\)

  2. \(A = \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx = \sqrt{2} \text{.}\)

  3. \(A = \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy = \frac{343}{48} \text{.}\)

  4. \(A = \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx = - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right) \text{.}\)

Solution

  1. The point of intersection of the graphs occurs when \(y(y - 2) = -(y - 1)(y - 3)\text{,}\) and thus we need to find where \(y^2 - 2y = -(y^2 - 4y + 3)\) or \(2y^2 - 6y + 3 = 0\text{.}\) By the quadratic formula, it follows that

    \begin{equation*} y = \frac{6 \pm \sqrt{(-6)^2 - (4)(2)(3)}}{(2)(2)} = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{3 \pm \sqrt{3}}{2}\text{.} \end{equation*}

    Since the curve \(x = -(y - 1)(y - 3)\) lies to the right (has a greater \(x\) value) of \(x = y(y - 2)\) between \(y = \frac{3 - \sqrt{3}}{2} \approx 0.63\) and \(y = \frac{3 + \sqrt{3}}{2} \approx 2.37\) as shown in the figure below, we see that the area between the two curves is given by the integral

    \begin{equation*} \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -(y - 1)(y - 3) - y(y - 2) \ dy\text{.} \end{equation*}

    Evaluating this integral, we find that

    \begin{align*} A &= \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy\\ &= \left. \left(-\frac{2}{3}y^3 + 3y^2 - 3y \right) \right|_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}}\\ &= \left[-\frac{2}{3} \left(\frac{3 + \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 + \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 + \sqrt{3}}{2}\right)\right]\\ &\qquad - \left[-\frac{2}{3} \left(\frac{3 - \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 - \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 - \sqrt{3}}{2}\right)\right]\\ &= -\frac{1}{12}(54+30\sqrt{3}) + \frac{1}{12}(54-30\sqrt{3}) + \frac{3}{4}(12+6\sqrt{3})\\ &\qquad - \frac{3}{4}(12-6\sqrt{3}) - \frac{3}{2}(3+\sqrt{3}) + \frac{3}{2}(3-\sqrt{3})\\ &= -5\sqrt{3} + 9\sqrt{3} - 3\sqrt{3}\\ &= \sqrt{3}\text{.} \end{align*}

  2. As the figure below shows, the graph of \(y = \sin(x)\) lies above the graph of \(y = \cos(x)\) on this interval. So the area between the two curves is given by

    \begin{align*} A &= \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx\\ &= \left. \left(-\cos(x) - \sin(x) \right) \right|_{\pi/4}^{3\pi/4}\\ &= \left[-\cos\left(\frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{4}\right) \right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) \right]\\ &= \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right] + \left[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right]\\ &= \sqrt{2}\text{.} \end{align*}

  3. We can rewrite \(y = 2x - 1\) as \(x = \frac{y+1}{2}\text{.}\) The two curves intersect when \(y^2 - y - 2 = \frac{y+1}{2}\text{.}\) It follows that \(2y^2 - 2y - 4 = y+1\text{,}\) so \(2y^2 - 3y - 5 = 0\) and thus factoring, \((2y-5)(y+1) = 0\text{.}\) Hence the two curves intersect when \(y = -1\) or \(y = \frac{5}{2}\text{.}\) The figure below shows that \(x = \frac{y+1}{2}\) lies to the right of \(x = y^2 - y - 2\)on this interval, so the area of the region between the two curves is

    \begin{align*} A &= \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy\\ &= \int_{-1}{5/2} -y^2 + \frac{3}{2}y + \frac{5}{2} \ dy\\ &= \left. \left( -\frac{y^3}{3} + \frac{3}{4}y^2 + \frac{5}{2}y \right) \right|_{-1}{5/2}\\ &= -\frac{1}{3}\left[\left(\frac{5}{2}\right)^3 + 1 \right] + \frac{3}{4}\left[\left(\frac{5}{2}\right)^2 - 1 \right] + \frac{5}{2}\left[\left(\frac{5}{2}\right) + 1 \right]\\ &= -\frac{133}{24} + \frac{63}{16} + \frac{35}{4}\\ &= \frac{343}{48}\text{.} \end{align*}

  4. A representative picture (with \(m = 2\)) is shown in the figure below. The two curves intersect when \(x^2 - 1 = mx\text{,}\) or when \(x^2 - mx - 1 = 0\text{.}\) By the quadratic formula, it folows that

    \begin{equation*} x = \frac{m \pm \sqrt{m^2-4(1)(-1)}}{2} = \frac{m \pm \sqrt{m^2+4}}{2}\text{.} \end{equation*}

    If we integrate with vertical slices, then the graph of \(y = mx\) is always above the graph of \(y = x^2 - 1\) and so the area of the region between the two graphs is

    \begin{align*} A &= \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx\\ &= \left. \left(-\frac{x^3}{3} + m\frac{x^2}{2} + x\right) \right|_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}}\\ &= - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2\\ &\qquad - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right)\text{.} \end{align*}
6.

Let \(f(x) = 1-x^2\) and \(g(x) = ax^2 - a\text{,}\) where \(a\) is an unknown positive real number. For what value(s) of \(a\) is the area between the curves \(f\) and \(g\) equal to 2?

Answer

\(a = \frac{1}{2}\text{.}\)

Solution

Both of these quadratic functions pass through the points \((-1,0)\) and \((1,0)\text{,}\) and thus these are the two intersection points for all positive values of \(a\text{.}\) In addition, since \(1 - x^2 \gt ax^2 - a\) on \(-1 \lt x \lt 1\text{,}\) it follows that the area between the curves is given by

\begin{equation*} A = \int_{-1}^{1} \left( 1 - x^2 - (ax^2 - a) \right) \, dx = \int_{-1}^{1} \left( (1+a) - (1+a)x^2 \right) \, dx\text{.} \end{equation*}

Evaluating the integral, we find that

\begin{align*} A &= \int_{-1}^{1} \left( 1 + a - (1+a)x^2 \right) \, dx\\ &= \left. (1+a)x - \frac{1+a}{3}x^3 \right|_{-1}^{1}\\ &= (1+a)(1) - \frac{1+a}{3}(1)^3 - \left( (1+a)(-1) - \frac{1+a}{3}(-1)^3 \right)\\ &= 2(1+a) - \frac{2(1+a)}{3}\\ &= \frac{4(1+a)}{3}\text{.} \end{align*}

We want to know for which value(s) of \(a\) it follows that \(A = 2\text{.}\) Thus, we solve

\begin{equation*} 2 = \frac{4(1+a)}{3}\text{,} \end{equation*}

from which it follows that \(a = \frac{1}{2}\text{.}\) Hence, in the figure below, the area between the two curves is exactly \(A = 2\text{.}\)

7.

Let \(f(x) = 2-x^2\text{.}\) Recall that the average value of any continuous function \(f\) on an interval \([a,b]\) is given by \(\frac{1}{b-a} \int_a^b f(x) \, dx\text{.}\)

  1. Find the average value of \(f(x) = 2-x^2\) on the interval \([0,\sqrt{2}]\text{.}\) Call this value \(r\text{.}\)

  2. Sketch a graph of \(y = f(x)\) and \(y = r\text{.}\) Find their intersection point(s).

  3. Show that on the interval \([0,\sqrt{2}]\text{,}\) the amount of area that lies below \(y = f(x)\) and above \(y = r\) is equal to the amount of area that lies below \(y = r\) and above \(y = f(x)\text{.}\)

  4. Will the result of (c) be true for any continuous function and its average value on any interval? Why?

Answer

  1. \(r = \frac{4}{3} \text{.}\)

  3. \(A_1 = A_2 = \frac{4 \sqrt{6}}{27}\text{.}\)

  4. Yes.

Solution

  1. The average value of \(f\) on the interval \([0,\sqrt{2}]\) is

    \begin{align*} r &= \frac{1}{\sqrt{2}} \int_0^{\sqrt{2}} 2-x^2 \, dx\\ &= \left. \frac{1}{\sqrt{2}}\left(2x - \frac{x^3}{3}\right) \right|_0^{\sqrt{2}}\\ &= \frac{1}{\sqrt{2}}\left( 2\sqrt{2} - \frac{\sqrt{2}^3}{3} \right)\\ &= 2 - \frac{2}{3}\\ &= \frac{4}{3}\text{.} \end{align*}

  2. The intersection point of \(y=f(x)\) and \(y=r\) occurs, for positive \(x\text{,}\) when \(2-x^2 = \frac{4}{3}\text{,}\) from which it follows that \(x^2 = \frac{2}{3}\) and thus \(x = \frac{\sqrt{6}}{3}\text{.}\) A graph of \(f\text{,}\) \(y = r\text{,}\) and the intersection point is shown in the following figure.

  3. Let \(A_1\) be the area that lies below \(y = f(x)\) and above \(y = r\) on the interval \(\left[0, \frac{\sqrt{6}}{3}\right]\) and \(A_2\) the are of the region that lies below \(y = r\) and above \(y = f(x)\) on \(\left[\frac{\sqrt{6}}{3}, \sqrt{2}\right]\) as illustrated in the figure below. Then

    \begin{align*} A_1 &= \int_0^{\sqrt{6}/3} (2-x^2) - \frac{4}{3} \ dx\\ &= \int_0^{\sqrt{6}/3} \frac{2}{3} - x^2 \ dx\\ &= \left. \left( \frac{2}{3}x - \frac{x^3}{3} \right) \right|_0^{\sqrt{6}/3}\\ &= \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) - \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3\\ &= \frac{2\sqrt{6}}{9} - \frac{6 \sqrt{6}}{81}\\ &= \frac{4 \sqrt{6}}{27}\text{.} \end{align*}

    and

    \begin{align*} A_2 &= \int_{\sqrt{6}/3}^{\sqrt{2}} \frac{4}{3} - (2-x^2) \ dx\\ &= \int_{\sqrt{6}/3}^{\sqrt{2}} x^2 - \frac{2}{3} \ dx\\ &= \left. \left(\frac{x^3}{3} - \frac{2}{3}x \right) \right|_{\sqrt{6}/3}^{\sqrt{2}}\\ &= \left(\frac{2\sqrt{2}}{3} - \frac{2}{3} \sqrt{2} \right) - \left[ \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 - \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) \right]\\ &= \frac{4 \sqrt{6}}{27}\text{.} \end{align*}

    Hence \(A_1 = A_2\) as desired.

  4. The answer is yes. Let \(AV\) is the average value of a function \(f\) on an interval \([a,b]\text{.}\) By the definition of average value,

    \begin{equation*} AV(b-a) = \int_a^b f(x) \ dx\text{,} \end{equation*}

    which can also be viewed as the area of the rectangle with base \((b-a)\) and height \(AV\text{.}\) Now suppose that the area of the region above the graph of \(f\) and below the line \(y=AV\) on the interval \([a,b]\) is \(R_1\) and the area of the region below the graph of \(f\) and above the line \(y=AV\) on the interval \([a,b]\) is \(R_2\text{.}\) The figure below illustrates that the value of the definite integral is the area of the rectangle plus the area above the rectangle minus the area below the rectangle, or

    \begin{equation*} \int_a^b f(x) \, dx = AV(b-a) + R_2 - R_1\text{.} \end{equation*}

    Having already established that \(AV(b-a) = \int_a^b f(x) \ dx\text{,}\) it follows that \(R_2 - R_1 = 0\text{.}\) So the area of the region that lies below \(y = f(x)\) and above \(y = AV\) is equal to the amount of area that lies below \(y = AV\) and above \(y = f(x)\) on \([a,b]\text{.}\)

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