Graph the integrand and use areas to evaluate the integral.
-$$\int _ { - 7 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x$$
A) $7 \pi$
B) $49 \pi$
C) 49
D) $\frac { 49 } { 2 } \pi$

Question

Quizplus · Accepted Answer

The integrand is the upper half of a circle with radius 7 centered at the origin.

Therefore, we can evaluate the integral by finding the area of half the circle and multiplying by 2:

$$\int _ { - 7 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x = 2 \int _ { 0 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x$$

Using the substitution $x = 7 \sin \theta$, we get:

$$\begin{aligned} 2 \int _ { 0 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x &= 2 \int _ { 0 } ^ { \frac{\pi}{2} } \sqrt { 49 - 49 \sin ^ 2 \theta } \cdot 7 \cos \theta d \theta \ &= 98 \int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta \ &= 98 \cdot \frac { \pi } { 4 } \ &= \boxed{\frac { 49 } { 2 } \pi}\end{aligned}$$

Graph the Integrand and Use Areas to Evaluate the Integral ∫−7749−x2dx\int _ { - 7 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x∫−77​49−x2​dx

Graph the Integrand and Use Areas to Evaluate the Integral $\int _ { - 7 } ^ { 7 } \sqrt { 49 - x ^ { 2 } } d x$