Question

selected values of the twice - differentiable function g are given in the table above. what is the value of $int_{0}^{4}g(x)arctan^{2}(2g(x)+3)dx$?
a -14.588
b -3.647
c -0.721
d 2.136

Explanation:

Step1: Use substitution

Let $u = 2g(x)+3$. Then $du=2g^{\prime}(x)dx$, and $g^{\prime}(x)dx=\frac{1}{2}du$.
When $x = 0$, $u=2g(0)+3=2\times1 + 3=5$.
When $x = 4$, $u=2g(4)+3=2\times(-\frac{5}{2})+3=-5 + 3=-2$.
So, $\int_{0}^{4}g^{\prime}(x)\arctan^{2}(2g(x)+3)dx=\frac{1}{2}\int_{5}^{-2}\arctan^{2}(u)du$.

Step2: Use integral property $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

$\frac{1}{2}\int_{5}^{-2}\arctan^{2}(u)du=-\frac{1}{2}\int_{-2}^{5}\arctan^{2}(u)du$.
We can use a calculator with integral - calculation function (such as a TI - 84 Plus: fnInt((atan(u))^2,u,-2,5)).
Evaluating $-\frac{1}{2}\int_{-2}^{5}\arctan^{2}(u)du\approx - 3.647$.

Answer:

B. $-3.647$