Home / AP Calculus AB : 8.6 Finding the Area Between Curves That Intersect at More Than Two Points- Exam Style questions with Answer- FRQ

AP Calculus AB : 8.6 Finding the Area Between Curves That Intersect at More Than Two Points- Exam Style questions with Answer- FRQ

Question:

Let R be the region in the first quadrant enclosed by the graphs of f(x) =  8x3 and g(x) = sin(πx) , p as shown in the figure above.
(a) Write an equation for the line tangent to the graph of f at \(x = \frac{1}{2}.\)
(b) Find the area of R.
(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the horizontal line y = 1.

▶️Answer/Explanation

Ans:

(a)

f(x) = 8x3               \(f\left ( \frac{1}{2} \right ) = 8\left ( \frac{1}{8} \right )=1\)        point of tangency : \(\left ( \frac{1}{2}, 1 \right )\)

f'(x) = 24x2          \(f’\left ( \frac{1}{2} \right )= 24 . \frac{1}{4}=6\)

y = 6x + b                                                              The tangent line to the graph of f

\(1 = 6\left ( \frac{1}{2} \right )+b\)          at \(x = \frac{1}{2}\)  is y = 6x-2.

1 = 3 + b

b = -2

y = 6x -2

(b)

f(x)  = g(x)

8×3 = sin(πx)

x = 0 ,  \(\frac{1}{2}\)

\(R = \int_{0}^{\frac{1}{2}}\left [ 9(x)-f(x) \right ]dx = \int_{0}^{\frac{1}{2}}\left [ sin(\pi x)-8x^{3} \right ]dx=\frac{1}{\pi }\int_{0}^{\frac{1}{2}}sinu du – \int_{0}^{\frac{1}{2}}8x^{3}dx\)

\(= \frac{1}{\pi }(-cos(\pi x))_{0}^{\frac{1}{2}}-2x^{4}\) \(_{0}^{\frac{1}{2}}\)

\(= \frac{1}{\pi }(-cos(\frac{\pi }{2})+cos(01))-2\left ( \frac{1}{2} \right )^{4}+0\)

\(= \frac{1}{\pi }(1)-\frac{2}{16}\)

\(= \frac{1}{\pi }-\frac{1}{8}=\frac{8-\pi }{8\pi }\)

(c)

\(\pi \int_{0}^{\frac{1}{2}}\left [ \left ( 1-f(x) \right )^{2}-\left ( 1-9(x) \right )^{2} \right ]dx\)

Question:

Let R be the region in the first quadrant bounded by the x-axis and the graphs of y = ln x and y = 5 – x, as shown in the figure above.
(a) Find the area of R.
(b) Region R is the base of a solid. For the solid, each cross section perpendicular to the x-axis is a square. Write, but do not evaluate, an expression involving one or more integrals that gives the volume of the solid.
(c) The horizontal line y = k divides R into two regions of equal area. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of k.

▶️Answer/Explanation

Ans:

(a)

ey = x        x = 5- y

1.307

\(A = \int_{0}^{1.307}(5-y-e^{y})dy\)

= 2.986

(b)

\(V = \int_{1}^{3.693}(In x)^{-2}dx +\int_{3.693}^{5}(5-x)^{-2}dx\)

(c)

\(V = \int_{0}^{k}(5-y-e^{y})dy =\int_{k}^{1.307}(5-y-e^{y})dy\)

Question:

Let \(f(x)= 2x^{2}-6x+4\) and \(g(x)= 4 cos \left ( \frac{1}{4}\pi x \right ).\) Let R be the region bounded by the graphs of f and g, as shown in the figure above.
(a) Find the area of R.
(b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 4.
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square.
Write, but do not evaluate, an integral expression that gives the volume of the solid. 

▶️Answer/Explanation

Ans:

(a)

\(R = \int_{0}^{2}\left ( 4 cos (\frac{1}{4}\pi x) -(2x^{2}-6x-4)\right )dx\)

\(R = \left ( \frac{16}{\pi } sin(\frac{1}{4}\pi x)-\frac{2}{3x^{3}}+3x^{2}-4x\right )_{0}^{2}\)

\(R = \left ( \frac{16}{\pi } sin(\frac{1}{2}\pi)-\frac{16}{3}+13-8\right )-\left ( \frac{16}{\pi } sin(0)-0+0-0\right )\)

\(R = \frac{16}{\pi }(1)-\frac{16}{3}+4\)

\(R = \frac{16}{\pi }-\frac{4}{3}\)

(b)

\(v = \pi \int_{0}^{2}\left ( 4-(2x^{2}-6x)+4 \right )^{2}-\left ( 4-4cos(\frac{1}{4}\pi x)\right )^{2}dx\)

\(v = \pi \int_{0}^{20}\left ( 2x^{2}+6x \right )^{2}-\left ( 4-4cos(\frac{1}{7}\pi x)\right )^{2}dx\)

(c)

\(v = \int_{0}^{2}\left ( 4cos(\frac{1}{4}\pi x)-( 2x^{2}-6x+4)\right )^{2}dx\)

\(v = \int_{0}^{2}\left ( 4cos(\frac{1}{4}\pi x)-( 2x^{2}+6x-4)\right )^{2}dx\)

Question:

Let R be the region enclosed by the graph of f(x) = x4 – 2.3x3 + 4 and the horizontal line y = 4, as shown in the figure above.
(a) Find the volume of the solid generated when R is rotated about the horizontal line y = -2.
(b) Region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in R. Find the volume of the solid.
(c) The vertical line x = k divides R into two regions with equal areas. Write, but do not solve, an equation involving integral expressions whose solution gives the value k.

▶️Answer/Explanation

Ans:

(a)

\(f(x)= 4| _{x = 0 or 2.3} \)               \(\pi \int_{0}^{2.3}\left ( (2+4)^{2}-(2+f(x))^{2} \right )dx = 98.868 units^{3}\)

\(V = \pi \left ( R^{2}-r^{2} \right )\)

(b)

\(A = \frac{1}{2}bh\)

\(A = \frac{1}{2}b^{2}\)

b = 4 – f(x)

\(\frac{1}{2}\int_{0}^{2.3}(4-p(x))^{2}dx= 3.574 units^{3}\)

(c)

\(\int_{0}^{k}(4-f(x))dx= \frac{1}{2}\int_{0}^{2.3}\left ( 4-f(x) \right )dx\)

Question:

Let R be the region enclosed by the graph of f(x) = x4 – 2.3x3 + 4 and the horizontal line y = 4, as shown in the figure above.
(a) Find the volume of the solid generated when R is rotated about the horizontal line y = -2.
(b) Region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in R. Find the volume of the solid.
(c) The vertical line x = k divides R into two regions with equal areas. Write, but do not solve, an equation involving integral expressions whose solution gives the value k.

▶️Answer/Explanation

Ans:

(a)

\(f(x)= 4| _{x = 0 or 2.3} \)               \(\pi \int_{0}^{2.3}\left ( (2+4)^{2}-(2+f(x))^{2} \right )dx = 98.868 units^{3}\)

\(V = \pi \left ( R^{2}-r^{2} \right )\)

(b)

\(A = \frac{1}{2}bh\)

\(A = \frac{1}{2}b^{2}\)

b = 4 – f(x)

\(\frac{1}{2}\int_{0}^{2.3}(4-p(x))^{2}dx= 3.574 units^{3}\)

(c)

\(\int_{0}^{k}(4-f(x))dx= \frac{1}{2}\int_{0}^{2.3}\left ( 4-f(x) \right )dx\)

Question:

Let f and g be the functions defined by \(f(x)= 1 + x + e^{x^{2}-2x}\) and g(x) = x4 – 6.5x2 + 6x + 2. Let R and S be the two regions enclosed by the graphs of f and g shown in the figure above.
(a) Find the sum of the areas of regions R and S.
(b) Region S is the base of a solid whose cross sections perpendicular to the x-axis are squares. Find the volume of the solid.
(c) Let h be the vertical distance between the graphs of f and g in region S. Find the rate at which h changes with respect to x when x = 1.8. 

▶️Answer/Explanation

Ans:

(a) 

f(a) = g(a)

a = 1.0328318883641

\(\int_{0}^{9}\left [ g(x)-f(x) \right ]dx+\int_{0}^{2}\left [ f(x)-g(x) \right ]dx=2.004\)

(b)

Cross sation area  A(x) = (f(x) – g(x))2              a = 1.0328318883641

\(\int_{2}^{9}A(x)dx=1.283\)

(c)

f'(1.8) = 2.1162821217136

g'(1.8) = 5.928

\(\frac{dh}{dx}=f'(1.8)-g'(1.8)=-3.812\)

Question:

Let R be the region enclosed by the graphs of \(g(x)= -2 + 3 cos \left ( \frac{\pi }{2}x \right )\)  and h(x) = 6 – 2(x-1)2, the y-axis, and the vertical line x = 2, as shown in the figure above.
(a) Find the area of R.
(b) Region R is the base of a solid. For the solid, at each x the cross section perpendicular to the x-axis has area \(A(x)=\frac{1}{x+3}.\) Find the volume of the solid.
(c) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 6.

▶️Answer/Explanation

Ans:

(a)

\(Area = \int_{0}^{2}h(x)-g(x)dx\)

\( = \int_{0}^{2}6-2(x-1)^{2}dx – \int_{0}^{2}-2+3 cos (\frac{\pi }{2}x)dx\)

\(=6x – \frac{2}{3}(x-1)^{3}|_{0}^{2}-\left ( -2x+\frac{6}{\pi }sin(\frac{\pi }{2}x) \right )|_{0}^{2}\)

\(=\left ( \left [ 12-\frac{2}{3}(1) \right ]-\left [ 0+\frac{2}{3} \right ] \right )-\left ( \left [ -4+\frac{6}{\pi }(0) \right ]-\left [ 0+0 \right ] \right )\)

\(=12-\frac{4}{3}+4=16-\frac{4}{3}\)

(b)

\(v = \int_{0}^{2}A(x)2x=\int_{0}^{2}\frac{1}{x+3}dx=ln|x+3||_{0}^{2}=ln |5| – ln |3|\)

= ln 5- ln 3

(c)

\(v = \int_{0}^{2}\pi \left [ (4-g(x))^{2}-(6-h(x))^{2} \right ]dx\)

Question:

Let R be the region enclosed by the graphs of \(g(x)= -2 + 3 cos \left ( \frac{\pi }{2}x \right )\)  and h(x) = 6 – 2(x-1)2, the y-axis, and the vertical line x = 2, as shown in the figure above.
(a) Find the area of R.
(b) Region R is the base of a solid. For the solid, at each x the cross section perpendicular to the x-axis has area \(A(x)=\frac{1}{x+3}.\) Find the volume of the solid.
(c) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 6.

▶️Answer/Explanation

Ans:

(a)

\(Area = \int_{0}^{2}h(x)-g(x)dx\)

\( = \int_{0}^{2}6-2(x-1)^{2}dx – \int_{0}^{2}-2+3 cos (\frac{\pi }{2}x)dx\)

\(=6x – \frac{2}{3}(x-1)^{3}|_{0}^{2}-\left ( -2x+\frac{6}{\pi }sin(\frac{\pi }{2}x) \right )|_{0}^{2}\)

\(=\left ( \left [ 12-\frac{2}{3}(1) \right ]-\left [ 0+\frac{2}{3} \right ] \right )-\left ( \left [ -4+\frac{6}{\pi }(0) \right ]-\left [ 0+0 \right ] \right )\)

\(=12-\frac{4}{3}+4=16-\frac{4}{3}\)

(b)

\(v = \int_{0}^{2}A(x)2x=\int_{0}^{2}\frac{1}{x+3}dx=ln|x+3||_{0}^{2}=ln |5| – ln |3|\)

= ln 5- ln 3

(c)

\(v = \int_{0}^{2}\pi \left [ (4-g(x))^{2}-(6-h(x))^{2} \right ]dx\)

Question

Let R and S in the figure above be defined as follows: R is the region in the first and second quadrants bounded by the graphs of 2  \(y = 3 -x^{2} \)and y =2^{x} .S is the shaded region in the first quadrant bounded by the two graphs, the x-axis, and the y-axis.
(a) Find the area of S.
(b) Find the volume of the solid generated when R is rotated about the horizontal line  y = -1.
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with one leg across the base of the solid. Write, but do not evaluate, an integral expression that gives the volume of the solid.

▶️Answer/Explanation

\(3-x^2=2^x\)  when \(x=-1.63658\) and x=1
let \(a=-1.636558\)
(A)area of S =\(\int_{0}^{1}2^xdx+\int_{1}^{\sqrt{3}}\left ( 3-x^2 \right )dx\)
=2.240
(B)Volume = \(\pi \int_{a}^{1}\left ( (3-x^2+1 \right )^2-\left ( (2^x+1)^2 \right )dx\)
=63.106 or 63.107
(C)Volume =\(\frac{1}{2}\int_{a}^{1}(3-x^2-2^x)dx\)

Scroll to Top