Estimate the intergral of $f(x)g'(x)$ from $0$ to $3$
By Andrew Adams •
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I am having a lot of trouble with this one, really can not get the correct solution.
2 Answers
$\begingroup$because of $(fg)'=fg'+gf'$ the integral is $$ f(3)g(3)-f(0)g(0)-\int_0^3 f'(x) g(x) dx = 36 - 2 \int_0^3 xg(x) dx \approx 36 - 2\left[ 0.5 \cdot 0 + 0.5 \cdot 0.5 \cdot 2.2 + \dots 0.5 \cdot 2.5 \cdot 3.3 \right] $$ or you use the right endpoints $$ \approx 36 - 2\left[ 0.5 \cdot 0.5 \cdot 2.2 + \dots 0.5 \cdot 3 \cdot 4 \right] $$
$\endgroup$ 4 $\begingroup$To estimate $\int_0^3 xg(x)\,dx$ you may apply Simpson's rule $$\frac{b-a}{6}\Bigl(f(a)+4f\bigl((a+b)/2\bigr)+f(b)\Bigr)$$ on the intervals $[0,1]$, $[1,2]$, and $[2,3]$ to get $$\int_0^3 xg(x)\,dx\approx\frac 16\bigl((0+4.4+2.5)+(2.5+18+6.2)+(6.2+31+12)\bigr)=12.3,$$ so the result should be $36-12.3=23.7$.
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