What is the integral from 0 to1 of $(\frac{4}{t^{2} + 1}) d t$ $(4/(t^2+1))dt$ ?

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Answer 1 · 2018-03-23T10:20:53

$\int_{0}^{1} (\frac{4}{t^{2} + 1}) d t = π$ $int_0^1(4/(t^2+1))dt=pi$

$\int_{0}^{1} (\frac{4}{t^{2} + 1}) d t = 4 \int_{0}^{1} (\frac{1}{1 + t^{2}}) d t$ $int_0^1(4/(t^2+1))dt=4int_0^1(1/(1+t^2))dt$

$= 4 {[arctan t]}_{0}^{1} = 4 (arctan 1 - arctan 0) = 4 (\frac{π}{4} - 0) = π$ $=4[arctant]_0^1=4(arctan1-arctan0)=4(pi/4-0)=pi$

What is the integral from 0 to1 of (4t2+1)dt(4/(t^2+1))dt?