Question

1. (5 points) if $f(0)=3$, $f(4)=2$, $f(0)=35$ and $f(4)= - 1$, and $f(x)$ is continuous, what is $int_{0}^{4}(5x + 1)f(x)dx$?

a. -36
b. -44
c. -51
d. -55
e. -79

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|_{a}^{b}-\int_{a}^{b}v\mathrm{d}u$. Let $u = 5x + 1$ and $\mathrm{d}v=f''(x)\mathrm{d}x$. Then $\mathrm{d}u = 5\mathrm{d}x$ and $v=f'(x)$.
So, $\int_{0}^{4}(5x + 1)f''(x)\mathrm{d}x=[(5x + 1)f'(x)]_{0}^{4}-\int_{0}^{4}5f'(x)\mathrm{d}x$.

Step2: Evaluate $[(5x + 1)f'(x)]_{0}^{4}$

$[(5x + 1)f'(x)]_{0}^{4}=(5\times4 + 1)f'(4)-(5\times0 + 1)f'(0)=(20 + 1)\times(-1)-1\times35=-21 - 35=-56$.

Step3: Evaluate $\int_{0}^{4}5f'(x)\mathrm{d}x$

We know that $\int_{0}^{4}5f'(x)\mathrm{d}x=5\int_{0}^{4}f'(x)\mathrm{d}x$. By the fundamental theorem of calculus, $5\int_{0}^{4}f'(x)\mathrm{d}x=5[f(x)]_{0}^{4}=5(f(4)-f(0))$.
Substitute $f(0) = 3$ and $f(4)=2$ into it, we get $5(2 - 3)=-5$.

Step4: Calculate the original integral

$\int_{0}^{4}(5x + 1)f''(x)\mathrm{d}x=-56-(-5)=-56 + 5=-51$.

Answer:

C. $-51$