selected values of the twice - differentiable function g are given in the table above. what is the value of…

Explanation:

Step1: Use substitution

Let $u = 2g(x)+1$. Then $du=2g^{\prime}(x)dx$, and $g^{\prime}(x)dx=\frac{1}{2}du$. When $x = 0$, $u=2g(0)+1=2\times5 + 1=11$. When $x = 3$, $u=2g(3)+1=2\times(-2)+1=-3$. So, $\int_{0}^{3}g^{\prime}(x)\cos^{2}(2g(x)+1)dx=\frac{1}{2}\int_{11}^{-3}\cos^{2}(u)du$. Since $\cos^{2}(u)=\frac{1 + \cos(2u)}{2}$, we have $\frac{1}{2}\int_{11}^{-3}\cos^{2}(u)du=\frac{1}{4}\int_{11}^{-3}(1+\cos(2u))du$.

Step2: Integrate term - by - term

$\frac{1}{4}\int_{11}^{-3}(1+\cos(2u))du=\frac{1}{4}\left[\int_{11}^{-3}1du+\int_{11}^{-3}\cos(2u)du\right]$. $\int_{11}^{-3}1du=u\big|{11}^{-3}=-3 - 11=-14$. For $\int{11}^{-3}\cos(2u)du$, let $v = 2u$, $dv = 2du$. Then $\int_{11}^{-3}\cos(2u)du=\frac{1}{2}\int_{22}^{-6}\cos(v)dv=\frac{1}{2}[\sin(v)]_{22}^{-6}=\frac{1}{2}(\sin(-6)-\sin(22))$. $\frac{1}{4}\left[-14+\frac{1}{2}(\sin(-6)-\sin(22))\right]$. Using a calculator, $\sin(-6)\approx0.2794$ and $\sin(22)\approx - 0.0088$. $\frac{1}{4}\left[-14+\frac{1}{2}(0.2794 + 0.0088)\right]=\frac{1}{4}\left[-14+\frac{1}{2}(0.2882)\right]=\frac{1}{4}\left[-14 + 0.1441\right]=\frac{-13.8559}{4}\approx - 3.464$.

Answer:

B. $-3.464$