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 method

Let $u = 2g(x)+3$, then $du=2g'(x)dx$, and $g'(x)dx=\frac{1}{2}du$.
When $x = 0$, $u=2g(0)+3=2\times1 + 3=5$.
When $x = 4$, $u=2g(4)+3=2\times(-\frac{5}{2})+3=-5 + 3=-2$.
The integral $\int_{0}^{4}g'(x)\arctan^{2}(2g(x)+3)dx$ becomes $\frac{1}{2}\int_{5}^{-2}\arctan^{2}(u)du$.

Step2: Use the property of definite - integral

$\frac{1}{2}\int_{5}^{-2}\arctan^{2}(u)du=-\frac{1}{2}\int_{-2}^{5}\arctan^{2}(u)du$.
We can use a calculator or software (such as a graphing calculator with integral - calculation function or Mathematica) to evaluate $\int_{-2}^{5}\arctan^{2}(u)du$.
$\int_{-2}^{5}\arctan^{2}(u)du\approx7.294$.
Then $-\frac{1}{2}\int_{-2}^{5}\arctan^{2}(u)du\approx - 3.647$.

Answer:

B. $-3.647$