Find the area enclosed by the given curves.
-Find the area of the region between the curve $y = 4 x / \left( 1 + x ^ { 2 } \right)$ and the interval $- 3 \leq x \leq 3$ of the $x$-axis.
A) 0
B) $\ln 10$
C) $4 \ln 10$
D) $4 \mathrm { e } ^ { 10 }$

Question

Quizplus · Accepted Answer

To find the area enclosed by the curve, we need to integrate the function from $x=-3$ to $x=3$.

$$A=\int_{-3}^{3} \frac{4x}{1+x^2} dx$$

We can solve this integral by substituting $u$ = 1+$x^2$. Then the integral becomes:

\begin{align*}
\int \frac{4x}{1+x^2} dx &=\ln|1+x^2|+C\
&=\ln(1+3^2)-\ln(1+(-3)^2) \
&= \ln(10)
\end{align*}

Therefore, the area enclosed by the curve is $\ln(10)$, which is option C.

Find the Area Enclosed by the Given Curves y=4x/(1+x2)y = 4 x / \left( 1 + x ^ { 2 } \right)y=4x/(1+x2)

Find the Area Enclosed by the Given Curves $y = 4 x / \left( 1 + x ^ { 2 } \right)$