$\text { Determine whether the improper integral } \int _ { 7 } ^ { 8 } \frac { 1 } { \sqrt { 64 - x ^ { 2 } } } d x \text { diverges or converges. }$ Evaluate the integral if it converges.
A) $\frac { \pi } { 2 } - \arcsin \left( \frac { 7 } { 8 } \right)$
B) $\frac { \pi } { 16 } - \arcsin \left( \frac { 7 } { 8 } \right)$
C) $\frac { \pi } { 2 } - \arcsin \left( \frac { 7 } { 64 } \right)$
D) $\frac { \pi } { 16 } - \arcsin \left( \frac { 7 } { 64 } \right)$
E) diverges

Question

$\text { Determine whether the improper integral } \int _ { 7 } ^ { 8 } \frac { 1 } { \sqrt { 64 - x ^ { 2 } } } d x \text { diverges or converges. }$ Evaluate the integral if it converges. 
A) $\frac { \pi } { 2 } - \arcsin \left( \frac { 7 } { 8 } \right)$
B) $\frac { \pi } { 16 } - \arcsin \left( \frac { 7 } { 8 } \right)$
C) $\frac { \pi } { 2 } - \arcsin \left( \frac { 7 } { 64 } \right)$
D) $\frac { \pi } { 16 } - \arcsin \left( \frac { 7 } { 64 } \right)$
E) diverges

Quizplus · Accepted Answer

The integral converges because the function $ \frac { 1 } { \sqrt { 64 - x ^ { 2 } } } $ is continuous on the interval [7,8], which is a subset of the domain of the function. The integral can be evaluated using a trigonometric substitution, $x = 8\sin(\theta)$, leading to the result $\frac { \pi } { 2 } - \arcsin \left( \frac { 7 } { 8 } \right)$.