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