How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{1} e^{- \sqrt{x}} d x$ $int_0^1e^-sqrt(x)dx$ ?

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{1} e^{- \sqrt{x}} d x$ $int_0^1e^-sqrt(x)dx$ ?

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Answer 1 · 2014-09-22T14:30:57

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

The definite integral

$\int_{0}^{1} f (x) d x$ $int_0^1 f(x) dx$

can be approximated by Trapezoid Rule

$\approx T_{6} = [f (\frac{0}{6}) + 2 f (\frac{1}{6}) + 2 f (\frac{2}{6}) + 2 f (\frac{3}{6}) + 2 f (\frac{4}{6}) + 2 f (\frac{5}{6}) + f (\frac{6}{6})] \frac{1 - 0}{2 (6)} \approx 0.54$ $approx T_6=[f(0/6)+2f(1/6)+2f(2/6)+2f(3/6)+2f(4/6)+2f(5/6)+f(6/6)]{1-0}/{2(6)}approx 0.54$

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{1} e^{- \sqrt{x}} d x$ $int_0^1e^-sqrt(x)dx$ ?

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{1} e^{- \sqrt{x}} d x$ $int_0^1e^-sqrt(x)dx$ ?