Solve the problem.
-Find the area bounded by $y \left( 36 + 4 x ^ { 2 } \right) = 2 , x = 0 , y = 0$, and $x = 4$.
A) $\sin ^ { - 1 } \left( \frac { 4 } { 3 } \right)$
B) $2 \sin ^ { - 1 } \left( \frac { 4 } { 3 } \right)$
C) $\frac { 1 } { 6 } \tan ^ { - 1 } \left( \frac { 2 } { 3 } \right)$
D) $\frac { 1 } { 6 } \tan ^ { - 1 } \left( \frac { 4 } { 3 } \right)$

Question

Quizplus · Accepted Answer

First, we need to solve the given equation for $y$. 
$$y=\frac{2}{36+4x^2}=\frac{1}{18+2x^2}$$
Now, we can find the area by integrating with respect to $x$. 
$$\int_{0}^{4} \frac{1}{18+2x^2} dx$$
Let $u=\sqrt{\frac{1}{9}x^2}$, $du = \frac{1}{3}x dx$
$$\int_{0}^{4} \frac{1}{18+2x^2} dx = \frac{1}{3}\int_{0}^{4} \frac{1}{1+u^2} du = \frac{1}{3} \tan^{-1}(u)|_0^4=\frac{1}{3}\tan^{-1}\left(\frac{4}{3}\right)$$
Thus, the answer is $\boxed{\text{(D)}}$.

Solve the Problem y(36+4x2)=2,x=0,y=0y \left( 36 + 4 x ^ { 2 } \right) = 2 , x = 0 , y = 0y(36+4x2)=2,x=0,y=0

Solve the Problem $y \left( 36 + 4 x ^ { 2 } \right) = 2 , x = 0 , y = 0$