Question

suppose that f(1)=4, f(4)=8, f(1)=6, f(4)=5, and f is continuous. find the value of $int_{1}^{4}xf(x)dx$.

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 $u = x$ and $\mathrm{d}v=f''(x)\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$ and $v = f'(x)$.
So, $\int_{1}^{4}xf''(x)\mathrm{d}x=[xf'(x)]_{1}^{4}-\int_{1}^{4}f'(x)\mathrm{d}x$.

Step2: Evaluate $[xf'(x)]_{1}^{4}$

$[xf'(x)]_{1}^{4}=4f'(4)-1\times f'(1)$. Substitute $f'(1) = 6$ and $f'(4)=5$ into it, we get $4\times5 - 1\times6=20 - 6 = 14$.

Step3: Evaluate $\int_{1}^{4}f'(x)\mathrm{d}x$

By the fundamental theorem of calculus, $\int_{1}^{4}f'(x)\mathrm{d}x=f(4)-f(1)$. Substitute $f(1) = 4$ and $f(4)=8$ into it, we get $8 - 4=4$.

Step4: Calculate the final result

$\int_{1}^{4}xf''(x)\mathrm{d}x=[xf'(x)]_{1}^{4}-\int_{1}^{4}f'(x)\mathrm{d}x=14-4 = 10$.

Answer:

$10$