4. Area and Average Value

Exercises

    a. Area Below a Curve

    In each of the following problems, find the area under the graph of \(f\), above the \(x\)-axis, between \(x=a\) and \(x=b\).

  1. \(f(x)=-x^2+4x-2\),   \(a=1\) and \(b=3\).

    This animation shows a parabola, bending down, in the first quadrant, with vertex above x equals 2. There are two vertial lines at x equals 1 and 3 and a collection of vertical lines between 1 and 3. There are gradually more and more vertial lines, until the region below the function between x equals 1 and 3 is totally filled.

    Remember the area under a function is the integral of the function: \[ A=\int_a^b f(x)\,dx \]

    \(A=\dfrac{10}{3}\)

    The area is \[\begin{aligned} A&=\int_1^3 (-x^2+4x-2)\,dx \\ &=\left[-\dfrac{x^3}{3}+2x^2-2x\right]_1^3 \\ &=(-9+18-6)-\left(-\dfrac{1}{3}+2-2\right) =\dfrac{10}{3} \end{aligned}\]

    jm,db 

  2. \(f(x)=x^3-2x\),   \(a=0\) and \(b=4\).

    \(\displaystyle A=48\)

    The area is \[\begin{aligned} A&=\int_0^4 (x^3-2x)\,dx =\left[\dfrac{x^4}{4}-x^2\right]_0^4 \\ &=\left[\dfrac{4^4}{4}-4^2\right]-0 =64-16=48 \end{aligned}\]

  3. \(f(x)=e^x+1\),   \(a=-1\) and \(b=2\).

    \(A=e^2-\dfrac{1}{e}+3\)

    The area is \[\begin{aligned} A&=\int_{-1}^{2} (e^x + 1)\,dx =\left[\dfrac{}{}e^x + x\right]_{-1}^{2} \\ &=\left[(e^2 + 2)-\left(\dfrac{1}{e}-1\right)\right] \\ &=e^2 -\dfrac{1}{e} + 3 \end{aligned}\]

    This plot shows an exponential function above the x interval from -1 to 2. There are two vertial lines at x equals -1 and 2. This plot is animated adding more and more vertial lines in the interval, until the region below the function is totally filled.

    db 

  4. \(f(x)=x(x^2+1)^{5}\),   \(a=0\) and \(b=1\).

    You'll need to use a substitution to do the integral.

    \(\displaystyle A=\dfrac{21}{4}\)

    The area is the integral: \[ A=\int_0^1 x(x^2+1)^{5}\,dx \] To do the integral, we make the substitution \(u=x^2+1\). Then \(du=2x\,dx\) and \(x\,dx=\dfrac{1}{2}\,du\). So the area is: \[ A=\dfrac{1}{2}\int_1^2 u^{5}du =\dfrac{1}{2}\left[\dfrac{u^{6}}{6}\right]_1^2 =\dfrac{2^6-1}{12} =\dfrac{21}{4} \]

    This plot shows an increasing function over the x interval from 0 and 1. It hits the x axis at x=0 and there is a vertical line at x=1. This plot is animated by adding more and more vertial lines between x equals 0 and x equals 1, until the region below the function is totally filled.

    jm,db 

  5. Challenge: The total sales of widgets sold since they went on the market is given by the function \(G(t)\) where \(t\) is time in days. So the rate they are being sold is \(G'(t)\). Explain in practical terms what the area of the region \(R\) represents where \(R\) is the region below the graph of \(G'(t)\), above the horizontal axis and between times \(t_1\)and \(t_2\).

    Recall the Fundamental Theorem of the Calculus.

    By the Fundamental Theorem: \[ \int_{t_1}^{t_2} G'(t)dt=G(t_2)-G(t_1) \] The left side of this equation is the area of \(R\). The right side is the difference between total sales by day \(t_2\) and by day \(t_1\). So the area represents the sales between day \(t_1\) and day \(t_2\).

    jm 

    6. b. Area Between Two Curves

    In each of the following problems, sketch the region between the graphs of \(f\) and \(g\) and find its area.

  6. \(f(x)=x^3\) and \(g(x)=x^2\)

    To find where they intersect, we equate them.
    To see which is on top, evaluate each function at \(x=\dfrac{1}{2}\).

    \(A=\dfrac{1}{12}\)

    This plot shows a concave up parabola with vertex at the origin, and a increasing cubic function with an infletion point at the origin switching from concave down to concave up. These two curves intersect at x equals 0 and 1 with the parabola above the cubic.

    To find where they intersect, we equate them: \[\begin{aligned} x^3=x^2 \quad &\Rightarrow \quad x^3-x^2=0 \\ &\Rightarrow \quad x^2(x-1)=0 \end{aligned}\] They intersect at \(x=0\) and \(x=1\).
    To see which is on top, we evaluate at \(x=\dfrac{1}{2}\): \[\begin{aligned} f\left(\dfrac{1}{2}\right)&=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8} \\ g\left(\dfrac{1}{2}\right)&=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4} \end{aligned}\]

    This plot shows a concave up parabola with vertex at the origin, and a increasing cubic function with an infletion point at the origin switching from concave down to concave up. These two curves intersect at x equals 0 and 1 with the parabola above the cubic.

    So \(g(x)=x^2\) is bigger and on top. So the area is: \[ A=\int_0^1 (x^2-x^3)\,dx =\left[\dfrac{x^3}{3}-\dfrac{x^4}{4}\right]_0^1 =\dfrac{1}{3}-\dfrac{1}{4} =\dfrac{1}{12} \]

    jm,db 

  7. \(f(x)=x-1\) and \(g(x)=x^2-4x+3\)

    \(A=\dfrac{9}{2}\)

    This plot shows a parabola opening upward with its vertex below the x axis and intersecting the x axis at x=1 and 3. Then there is a line with an increasing slope, intersecting with the parabola at x equals 1 and 4.

    We first find the points of intersection: \[\begin{aligned} x-1&=x^2-4x+3 \\ 0&=x^2-5x+4=(x-4)(x-1) \\ &\Rightarrow x=1,4 \end{aligned}\] Since \(f\) is the upper function, we evaluate: \[\begin{aligned} \int_1^4 &[(x-1)-(x^2-4x+3)]\,dx \\ &=\int_1^4 (-x^2+5x-4)\,dx \\ &=\left[-\dfrac{x^3}{3}+\dfrac{5x^2}{2}-4x\right]_1^4 \\ &=-\dfrac{64}{3}+40-16+\dfrac{1}{3}-\dfrac{5}{2}+4 =\dfrac{9}{2} \end{aligned}\]

    This plot shows a parabola opening upward with its vertex below the x axis and intersecting the x axis at x=1 and 3. Then there is a line with an increasing slope, intersecting with the parabola at x equals 1 and 4.

    jm,db 

  8. \(f(x)=x^3-4x\) and \(g(x)=5x\)

    \(A=\dfrac{81}{2}\)

    This plot shows a cubic function with an infletion point at the origin, changing from concave down to concave up. There is also a line with increasing slope which crosses the cubic function at x equals negative 3 (in the third quadrant), the origin, and positive 3 (in the first quadrant).

    We first solve for the intersection points: \[\begin{aligned} &x^3-4x=5x \quad \Rightarrow \quad x^3-9x=0 \\ &\Rightarrow \quad x(x-3)(x+3)=0 \quad \Rightarrow \quad x=-3,0,3 \end{aligned}\] So we need to break the integral into two pieces. From the graph we see that the cubic is on top for the first integral while the line is on top for the second integral. We can check this by plugging in \(x=1\): \[ f(1)=-3 \qquad g(1)=5 \] So \(g(x)\) is bigger on the right. The area is:

    This plot shows a cubic function with an infletion point at the origin, changing from concave down to concave up. There is also a line with increasing slope which crosses the cubic function at x equals negative 3 (in the third quadrant), the origin, and positive 3 (in the first quadrant).

    \[\begin{aligned} A&=\int_{-3}^0 \text{upper}-\text{lower}\,dx +\int_0^3 \text{upper}-\text{lower}\,dx \\ &=\int_{-3}^0 [(x^3-4x)-5x]\,dx+\int_0^3 [5x-(x^3-4x)]\,dx \\ &=\int_{-3}^0 [x^3-9x]\,dx+\int_0^3 [9x-x^3]\,dx \\ &=\left[\dfrac{x^4}{4}-9\dfrac{x^2}{2}\right]_{-3}^0 +\left[ 9\dfrac{x^2}{2}-\dfrac{x^4}{4}\right]_0^3 \\ &=[ 0]-\left[\dfrac{81}{4}-9\dfrac{9}{2}\right] +\left[ 9\dfrac{9}{2}-\dfrac{81}{4}\right]-[ 0] =\dfrac{81}{2} \end{aligned}\]

    jm 

    Since the region is symmetric, we could have doubled the area of one piece: \[ A=2\int_0^3 [ 5x-(x^3-4x)]\,dx \]

    9. In each of the following problems, sketch the region bounded by the given curves and find its area.

  9. \(y=e^x\), \(y=e^{-x}\) and \(x=2\)

    Notice this region is to the right of both exponentials.

    \(A=e^2+e^{-2}-2\)

    This plot shows an exponential function e to the x on the top, and an exponential function e to the negative x on the bottom above the x interval from 0 to 2. There is a vertial line at x equals 2, forming an 'triangular' area between the curves and x = 2.

    The exponentials intersect when \(e^x=e^{-x}\) or \(e^{2x}=1\) or \(x=0\). So the area is: \[\begin{aligned} A&=\int_0^2 (e^x-e^{-x})\,dx =[e^x+e^{-x}]_0^2 \\ &=[e^2+e^{-2}]-[e^0+e^0] =e^2+e^{-2}-2 \end{aligned}\]

    This plot shows an exponential function e to the x on the top, and an exponential function e to the negative x on the bottom above the x interval from 0 to 2. There is a vertial line at x equals 2, forming an 'triangular' area between the curves and x = 2.

    jm,db 

  10. \(y=e^x\), \(y=e^{-x}\) and \(y=2\)

    Notice this region is above both exponentials and is symmetric about the \(y\)-axis.

    \(A=4\ln 2-2\)

    This plot shows two exponential functions e to the x which is increasing and e to the negative x which is decreasing. They cross on the y axis at y=1. There is also a horizontal line at y equals 2 which intersects with the exponentials at x = minus natural log 2 and x = positive natural log 2. The 3 curves form a 'triangular' region above the two exponentials and below the line.

    By symmetry, we will double the area to the right of \(x=0\). The line \(y=2\) intersects the exponential \(y=e^x\) at \(x=\ln2\). So: \[\begin{aligned} A&=2\int_0^{\ln 2} (2-e^x)dx =2\left[\dfrac{}{}2x-e^x\right]_0^{\ln 2} \\ &=2[ 2\ln 2-e^{\ln 2}-(0-1)] \\ &=2[ 2\ln 2-2+1] =4\ln 2-2 \end{aligned}\]

    This plot shows two exponential functions e to the x which is increasing and e to the negative x which is decreasing. They cross on the y axis at y=1. There is also a horizontal line at y equals 2 which intersects with the exponentials at x = minus natural log 2 and x = positive natural log 2. The 3 curves form a 'triangular' region above the two exponentials and below the line.

    jm,db 

    11. In each problem, sketch the region(s) bounded by the graph of \(f\) and the \(x\)-axis. Then compute its area.

  11. \(f(x)=x^3+x^2-6x\)

    This is the area between 2 curves, one the graph of \(f\) and the other the \(x\)-axis.

    \(A=\dfrac{253}{12}\)

    This plot shows a cubic function which passes through the origin, concave down on the left and concave up on the right. The curve intersects the x axis at negative 3 and positive 2. The area between the curve and x axis is shaded and is above the the x axis in the second quadrant and below the x-axis in the fourth quadrant.

    We first find x-intercepts: \[ x^3+x^2-6x=x(x-2)(x+3)=0 \] \[ x=-3,0,2 \] From the factored formula, we see that the cubic is positive for \(-3 \lt x \lt 0\) and negative for \(0 \lt x \lt 2\). So the cubic is on top in the first integral and on bottom in the second integral:

    This plot shows a cubic function which passes through the origin, concave down on the left and concave up on the right. The curve intersects the x axis at negative 3 and positive 2. The area between the curve and x axis is shaded and is above the the x axis in the second quadrant and below the x-axis in the fourth quadrant.

    \[\begin{aligned} A&=\int_{-3}^0 \text{upper}-\text{lower}\,dx +\int_0^2 \text{upper}-\text{lower}\,dx \\ &=\int_{-3}^0 [(x^3+x^2-6x)-0]\,dx+\int_0^2 [0-(x^3+x^2-6x)]\,dx \\ &=\dfrac{63}{4}+\dfrac{16}{3}=\dfrac{253}{12} \end{aligned}\]

    jm,db 

    12. c. Area as a y-Integral

    In each problem, sketch the region(s) bounded by the graph of \(f\) and the \(y\)-axis. Then compute its area.

  12. \(x=\sin y\)   for \(0 \le y \le \pi\)

    \(A=2\)

    The area is: \[\begin{aligned} A&=\int_0^\pi\sin y\,dy =\left[-\cos y\rule{0pt}{12pt}\right]_0^\pi \\ &=-(-1)-(-1)=2 \end{aligned}\]

    This plot shows graph of x equals sine of y, with a vertical line along the y axis between 0 and pi. The area between the curve and the y axis is shaded.

    jm,db 

  13. \(x=16-y^4\)

    \(A=\dfrac{256}{5}\)

    We find the endpoints: \[ 16-y^4=0 \quad \Rightarrow \quad y=\pm2 \] \[\begin{aligned} A&=\int_{-2}^2 (16-y^4)\,dy =\left[16y-\dfrac{y^5}{5}\right]_{-2}^2 \\ &=2\left(32-\dfrac{32}{5}\right) =\dfrac{256}{5} \end{aligned}\]

    This plot shows a curve similar to a parabola opening to the left with vertex at on the x axis at x equals 16. It intersects the y axis at y equals 2 and negative 2, forming an area between the curve and the y axis.

    jm,db 

    14. In each of the following problems, sketch the region bounded by the graphs of the given functions and find its area.

  14. \(y^2=1+x\) and the line joining \((-1,0)\) and \((0,1)\)

    Write the edges with \(x\) as a function of \(y\). What are the endpoints? Which curve is on the right (larger \(x\))?

    \(A=\dfrac{1}{6}\)

    As functions of \(y\) the edges are the parabola \(x=y^2-1\) and the line \(x=y-1\). From the plot we see they intersect at \(y=0\) and \(y=1\). However, if we do not have a plot, we can find where they intersect by equating them: \[ y^2-1=y-1 \quad \text{or} \quad y=0,1 \] From the plot we see that the line is on the right and the parabola is on the left. However, if we do not have a plot, we can plug in \(y=\dfrac{1}{2}\) into the parabola to get \(x=-\dfrac{3}{4}\) and into the line to get \(x=-\dfrac{1}{2}\). So the line is on the right.

    This plot shows the upper half of a parabola with x as a function of y opening to the right and vertex at x equals minus 1. The plot also shows a line with positive slope, intersecting the parabola at the vertex and where it crosses the y axis at y = 1. The region below the curve above the line is shaded.

    Then the area is: \[\begin{aligned} A&=\int_0^1 (y-1)-(y^2-1)\,dy =\int_0^1 (y-y^2)\,dy \\ &=\left[\dfrac{y^2}{2}-\dfrac{y^3}{3}\right]_0^1 =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6} \end{aligned}\]

    jm,db 

  15. \(x=y^3-9y\) and the \(y\)-axis.

    What are the intersection points? When is \(x \gt 0\)? When \(x \lt 0\)?

    \(A=\dfrac{81}{2}\).

    The function intersect the \(y\)-axis when \(0=y^3-9y=y(y^2-9)\) or \(y=-3,0,3\).
    When we plug in \(y=1\), we get \(x=-8\).
    When we plug in \(y=-1\), we get \(x=8\).
    So the curve is to the left of the \(y\)-axis for \(y \ge 0\) and to the right of the \(y\)-axis for \(y \le 0\).

    The plot shows a cubic function with x as a function of y. It has an inflection point at the origin, opening to the right above the x axis and to the left below the x axis. It intersect y axis at y equals 3, 0 and negative 3. Forming an area between the curve and the y axis in the second and fourth quadrants.

    So the area is: \[\begin{aligned} A&=\int_{-3}^0 (y^3-9y)-(0)\,dy+\int_0^3 (0)-(y^3-9y)\,dy \\ &=\left[\dfrac{y^4}{4}-\dfrac{9y^2}{2}\right]_{-3}^0 +\left[-\dfrac{y^4}{4}+\dfrac{9y^2}{2}\right]_0^3 \\ &=2\left(-\dfrac{81}{4}+\dfrac{81}{2}\right) =\dfrac{81}{2} \end{aligned}\]

    jm 

    16. d. Average Value and Mean Value Theorem

    In each of the following problems, find the average value of the given function on the given interval. Plot the function and its average.

  16. \(f(x)=x^3\) on \([-1,2]\)

    This plot shows an increasing cubic function with a horizontal inflection point at the origin which is concave down on the left and concave up on the right. The curve is shown over the x interval from minus 1 to 2 and between minus 1 and 8 on the y axis.

    \(f_\text{ave}=\dfrac{5}{4}\)

    This plot shows the same cubic as above but also shows a horizontal line at y equals 5 quarters which is the average of this function.

    \[\begin{aligned} f_\text{ave} &=\dfrac{1}{2-(-1)}\int_{-1}^2 x^3\,dx =\dfrac{1}{3}\left[ \dfrac{x^4}{4}\right]_{-1}^2 \\ &=\dfrac{1}{12}(16-1) =\dfrac{5}{4} \end{aligned}\]

    This plot shows the same cubic as above but also shows a horizontal line at y equals 5 quarters which is the average of this function.

    jm,py 

  17. \(f(x)=x^4-x^2\) on \([ 0,2]\)

    This plot shows a function similar to a parabola opening upward above in the x interval from 0 and 2. It passes through the origin and x equals 1

    \(f_\text{ave}=\dfrac{28}{15}\)

    This plot shows the same curve as above but also shows a horizontal line at y equals 28 over 15 which is the average of this function.

    \[\begin{aligned} f_\text{ave} &=\dfrac{1}{2}\int_0^2 (x^4-x^2)\,dx =\dfrac{1}{2}\left[\dfrac{x^5}{5}-\dfrac{x^3}{3}\right]_0^2 \\ &=\dfrac{1}{2}\left(\dfrac{32}{5}-\dfrac{8}{3}\right)=\dfrac{28}{15} \end{aligned}\]

    This plot shows the same curve as above but also shows a horizontal line at y equals 28 over 15 which is the average of this function.

    jm,py 

  18. The identity \(\sin^2 x=\dfrac{1-\cos(2x)}{2}\) allows us to do the following integrals. See the chapter on Trig Integrals.

    1. Compute \(\displaystyle \int \sin^2 x\,dx\).

      \(\displaystyle \int \sin^2 x\,dx =\dfrac{1}{2}\left(x-\dfrac{\sin(2x)}{2}\right)+C \)

      \[\begin{aligned} \int \sin^2 x\,dx &=\dfrac{1}{2}\int [1-\cos(2x)]\,dx \\ &=\dfrac{1}{2}\left(x-\dfrac{\sin(2x)}{2}\right)+C \end{aligned}\]

      jm 

      If   \(f(x)=\dfrac{1}{2}\left(x-\dfrac{\sin(2x)}{2}\right)\) then: \[ \dfrac{df}{dx} =\dfrac{1}{2}\left(1-\cos(2x)\right) =\sin^2 x \]

    2. Find the average value of \(\sin^2 x\) on \(\left[0,\dfrac{\pi}{4}\right]\).

      The average value of \(f\) on \(\left[0,\dfrac{\pi}{4}\right]\) is \(\displaystyle f_\text{ave}=\dfrac{4}{\pi}\int_0^{\pi/4} f(x)\,dx\).

      The average value of \(\sin^2 x\) on \(\left[0,\dfrac{\pi}{4}\right]\) is \(\dfrac{1}{2}-\dfrac{1}{\pi}\).

      If \(f=\sin^2 x\), then \[\begin{aligned} f_\text{ave} &=\dfrac{4}{\pi}\int_0^{\pi/4} \sin^2 x\,dx \\ &=\dfrac{4}{\pi}\dfrac{1}{2}\left[x-\dfrac{\sin(2x)}{2}\right]_0^{\pi/4} \\ &=\dfrac{2}{\pi}\left(\dfrac{\pi}{4}-\dfrac{\sin\dfrac{\pi}{2}}{2}\right) =\dfrac{1}{2}-\dfrac{1}{\pi} \end{aligned}\]

      jm 

    3. Find the average value of \(\sin^2 x\) on \([0,2\pi]\).

      The average value of \(f\) on \([0,2\pi]\) is \(\displaystyle f_\text{ave}=\dfrac{1}{2\pi}\int_0^{2\pi} f(x)\,dx\).

      The average value of \(\sin^2 x\) on \([0,2\pi]\) is \(\dfrac{1}{2}\).

      If \(f=\sin^2 x\), then \[\begin{aligned} f_\text{ave} &=\dfrac{1}{2\pi}\int_0^{2\pi} \sin^2 x\,dx \\ &=\dfrac{1}{2\pi}\dfrac{1}{2}\left[x-\dfrac{\sin(2x)}{2}\right]_0^{2\pi} \\ &=\dfrac{1}{4\pi}\left(2\pi-\dfrac{\sin4\pi}{2}\right) =\dfrac{1}{2} \end{aligned}\]

      jm 

      This should not be surprising. If we look at the graph of \[ y=\sin^2 x=\dfrac{1-\cos(2x)}{2} \] the center line is \(y=\dfrac{1}{2}\).

      This plot shows a graph of sine squared of x in the first quadrant. Since this function equals one half of 1 minus cos of 2 x, the curve wiggles between 0 and 1 with an average value of one half. There is a dashed horizontal line at this average value.

      py 

    19. In each of the following problems, find the height of a rectangle whose base is the given interval on the \(x\)-axis and whose area matches that of the region below the given function on the given interval. Plot the function and the rectangle.

  19. \(y=e^x\) on \([0,2]\)

    The height is the average value of the function.

    \(y_\text{ave}=\dfrac{e^2-1}{2} \approx3.2\)

    This plot shows a exponential function above the x interval from 0 to 2, and a green rectangle with the samr based with height at y equals 3.2 which is the average of this function.

    \[\begin{aligned} y_\text{ave} &=\dfrac{1}{2}\int_0^2 e^x\,dx =\dfrac{1}{2}\left[e^x\rule{0pt}{12pt}\right]_0^2 \\ &=\dfrac{e^2-1}{2} \approx3.2 \end{aligned}\] Notice the area of the rectangle is the same as the area below the curve.

    This plot shows a exponential function above the x interval from 0 to 2, and a green rectangle with the samr based with height at y equals 3.2 which is the average of this function.

    jm,py 

  20. \(f(x)=x^4-4x^2+4\) on \([-2,2]\)

    The height is the average value of the function.

    \(f_\text{ave}=\dfrac{28}{15}\)

    This plot shows a W shaped curve above the x interval from negative 2 to 2, and a shaded rectangle with the same base with height at y equals 28 over 15 which is the average of this function.

    \[\begin{aligned} f_\text{ave} &=\dfrac{1}{2-(-2)}\int_{-2}^2 (x^4-4x^2+4)\,dx \\ &=\dfrac{1}{4}\left[\dfrac{x^5}{5}-\dfrac{4x^3}{3}+4x\right]_{-2}^2 \\ &=2\cdot\dfrac{1}{4}\left(\dfrac{2^5}{5}-\dfrac{4\cdot2^3}{3}+4\cdot2\right) \\ &=\dfrac{2^4}{5}-\dfrac{4\cdot2^2}{3}+4 \\ &=\dfrac{48-80+60}{15}=\dfrac{28}{15} \end{aligned}\]

    This plot shows a W shaped curve above the x interval from negative 2 to 2, and a shaded rectangle with the same base with height at y equals 28 over 15 which is the average of this function.

    jm,py 

    21. In each of the following problems, find the value(s) of \(c\) guaranteed by the Mean Value Theorem for Integrals for the given function on the given interval.

  21. \(f(x)=e^x\) on \([ 0,2]\)

    Solve \(f(c)=f_\text{ave}\).

    \(c=\ln\dfrac{e^2-1}{2}\approx1.16\)

    Since \(f_\text{ave}=\dfrac{e^2-1}{2}\), (See a previous exercise.) we solve \[ f(c)=e^c=f_\text{ave}=\dfrac{e^2-1}{2} \] to get: \[ c=\ln\dfrac{e^2-1}{2}\approx1.16 \]

    This plot shows a exponential function above the x interval from 0 to 2, and a shaded rectangle with the same base with height y = 3.2 which is the average of this function. The plot also shows a vertial line at x equals 1.16 from the x axis to the curve at y = 3.2.

    jm,py 

  22. \(f(x)=x^4\) on \([ 0,5]\)

    Solve \(f(c)=f_\text{ave}\).

    \(c=5^{3/4}\approx3.34\)

    We first find the average: \[ f_\text{ave} =\dfrac{1}{5}\int_0^5 x^4\,dx =\left[\dfrac{1}{25}x^5\right]_0^5 =125 \] Then we solve \(f(c)=f_\text{ave}\): \[ c^4=125 \quad \Rightarrow \quad c=5^{3/4}\approx3.34 \]

    This plot shows the quartic function y = x to the fourth above the x interval from 0 to 5. There is also a shaded rectangle with the same base with height y equals 125 which is the average of this function. The plot also shows a vertical line from the x axis at x = 3.34 to the curve at y = 125.

    jm 

    23. e. Tutorial on Average Value

    Do the tutorial repeatedly until you feel confident in estimating and finding averages.

    Review Exercises

  23. Find the area of the region between \(y=\cos x\) and \(y=1\), for \(0 \le x \le \pi\).

    \(A=\pi\)

    \[ A=\int_0^\pi 1-\cos x\,dx =\left[x-\sin x\right]_0^\pi =\pi-\sin\pi =\pi \]

    jm 

    This should not be surprising. If we look at the graph of \(\cos x\), we see the required area is half of the area of the rectangle, \(0 \le x \le \pi\) and \(-1 \le y \le 1\), whose area is \(2\pi\).

    This plot shows the graph of cosine, on the x interval from 0 to pi, and the y interval from negative 1 to 1. The curve splits the area in half.

    py 

    24. Find the area of the region bounded by the given curves and lines.

  24. \(x=1-y^{2}\), \(y\)-axis

    Is this an \(x\)-integral or a \(y\)-integral?

    \(A = \dfrac{4}{3}\)

    The curve \(x=1-y^{2}\) intersects the \(y\)-axis at \(y=\pm1\). So the area is: \[\begin{aligned} A&=\int_{-1}^1 (1-y^2) \,dy =\left[y-\dfrac{y^{3}}{3}\right]_{-1}^1 \\ &=1-\dfrac{1^3}{3}-\left(-1-\dfrac{(-1)^3}{3}\right) =2(1-\dfrac{1}{3}) =\dfrac{4}{3} \end{aligned}\]

    jr 

  25. \(y=x^{3}-3x^{2}\), \(x\)-axis

    Where do the function intersect? Which function is bigger in each piece?

    \(A = \dfrac{27}{4}\)

    Let \(f(x)=x^{3}-3x^{2}\) and \(g(x)=0\). We first find where the curves intersect: \[ x^{3}-3x^{2}=0 \implies x=0,3 \] To see which is bigger, we try a point in between, \(x=1\): \[ f(1)=1^{3}-3\cdot1^{2}=-2 \quad\text{and}\quad g(1)=0 \] So \(g(x)=0\) is on top. This is confirmed in a plot.

    This plot shows a cubic function with a local maximum at the origin, and a local minimum at x = 2 and y = negative 4. It intersects the x axis at x equals 3. The area above the curve below the x axis for x between 0 and 3 is shaded.

    So the area is: \[\begin{aligned} A&=\int_0^3 g(x)-f(x)\,dx =\int_0^3 -x^3+3x^2 \,dx \\ &=\left[-\,\dfrac{x^4}{4}+x^3\right]_0^3 =-\,\dfrac{81}{4}+27 =-\,\dfrac{81}{4}+\dfrac{108}{4} =\dfrac{27}{4} \end{aligned}\]

    jr 

  26. \(y=x^{3}-3x\), \(x\)-axis

    Where do the function intersect? Which function is bigger in each piece?

    \(A = \dfrac{9}{2}\)

    Let \(f(x)=x^{3}-3x\) and \(g(x)=0\). We first find where the curves intersect: \[ x^{3}-3x=0 \implies x=0,\pm\sqrt 3 \] We find which is on top or bottom by evaluating \(f\) at a number in each interval: \[\begin{aligned} f(-1)&=-1+3=2 \gt 0 \\ f(1)&=1-3=-2 \lt 0 \end{aligned}\] So the \(f\) is on the top from \(-\sqrt3\) to \(0\) and on the bottom from \(0\) to \(\sqrt3\). This is confirmed in a plot.

    This plot shows a cubic function on the x interval from minus 2 to 2. It has a local maximum y = 2 at x = negative 1 and a local minimum y = minus 2 at x equals 1. The function intersects x axis at the origin and at x equals plus and minus square root of 3. The area between the curve and x axis is shaded in two pieces.

    So the area is: \[\begin{aligned} A&=\int_{-\sqrt 3}^0 (f(x)-g(x))\,dx+\int_0^{\sqrt 3} (g(x)-f(x))\,dx \\ &=\int_{-\sqrt 3}^0 (x^3-3x)\,dx+\int_0^{\sqrt 3} (-x^3+3x)\,dx \\ &=\left[\dfrac{x^4}{4}-\dfrac{3x^2}{2}\right]_{-\sqrt 3}^0 +\left[-\,\dfrac{x^4}{4}+\dfrac{3x^2}{2}\right]_0^{\sqrt 3} \\ &=-\left(\dfrac{9}{4}-\dfrac{9}{2}\right)+\left(-\dfrac{9}{4}+\dfrac{9}{2}\right) =2\left(\dfrac{9}{4}\right) = \dfrac{9}{2} \end{aligned}\] Note: We could have doubled either one of these integrals since the region is symmetric.

    jr 

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