## Calculus: Early Transcendentals 8th Edition

a) A $\approx 0.38$ b) V$\approx 0.87$
{Step 1 of 4} A) Divide the interval [0,1] into 4 sub-intervals (n=4). The width of each sub-interval $\Delta x=\frac{1}{4}$ The sub-intervals are $\left[ 0,\frac{1}{4} \right] , \left[ \frac{1}{4},\frac{1}{2} \right] , \left[ \frac{1}{2},\frac{3}{4} \right] , and \left[ \frac{3}{4},1 \right] .$ Their midpoints are $\frac{1}{8},\frac{3}{8},\frac{5}{8}, and \frac{7}{8}$. {Step 2 of 4} Use the midpoint rule to estimate the area of region R as A=$\int _{0}^{1}tan \left( x^{2} \right) dx$ $\approx \frac{1}{4} \left[ tan \left( \left( \frac{1}{8} \right) ^{2} \right) +tan \left( \left( \frac{3}{8} \right) ^{2} \right) +tan \left( \left( \frac{5}{8} \right) ^{2} \right) +tan \left( \left( \frac{7}{8} \right) ^{2} \right) \right]$ $\approx \frac{1}{4} \left[ 0.015626+0.141559+0.411786+0.961216 \right]$ $\approx \frac{1}{4} \left[ 1.530187 \right]$ A $\approx 0.38$ (up to 2 decimal places) {Step 3 of 4} B) Use the midpoint rule to find the volume of the solid generated by rotating region R about the x-axis. This produces a disk of radius $f(x)=tan(x^2)$ The area of the disk A(x) = $\pi tan^{2} \left( x^{2} \right)$ {Step 4 of 4} Then volume V $\approx \sum _{i=1}^{4}A \left( x_{i}^{.} \right) . \Delta x \approx \frac{ \pi }{4} \left[ tan^{2} \left( \left( \frac{1}{8} \right) ^{2} \right) +tan^{2} \left( \left( \frac{3}{8} \right) ^{2} \right) +tan^{2} \left( \left( \frac{5}{8} \right) ^{2} \right) +tan^{2} \left( \left( \frac{7}{8} \right) ^{2} \right) \right]$ $\approx \frac{ \pi }{4} \left[ 0.000+0.020+0.924 \right]$ $\approx \frac{ \pi }{4} \left( 1.114 \right)$ V$\approx 0.87$