How do you Use the trapezoidal rule with $n = 10$ $n=10$ to approximate the integral $\int_{0}^{2} \sqrt{x} \cdot e^{- x} d x$ $int_0^2sqrt(x)*e^(-x)dx$ ?

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Answer 1 · 2014-11-14T23:28:26

Let $f (x) = \sqrt{x} e^{- x} = \frac{\sqrt{x}}{e^{x}}$ $f(x)=sqrt{x}e^{-x}=sqrt{x}/e^x$ .

$Delta x={b-a}/n={2-0}/10=0.2$

By summing up the areas of trapezoids shown below,

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we have

$T_{10}=[f(0)+2f(0.2)+2f(0.4)+cdots+2f(1.8)+f(2)]{Delta x}/2approx 0.64$

I hope that this was helpful.

How do you Use the trapezoidal rule with n=10n=10n=10 to approximate the integral int_0^2sqrt(x)*e^(-x)dx∫20√x⋅e−xdxint_0^2sqrt(x)*e^(-x)dx?