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Evaluate the following definite integrals: (a) 0xexdx\int_0^{\infty} x e^{-x} d x (b) 01[(x3+1)/(x4+4x+1)]dx\int_0^1\left[\left(x^3+1\right) /\left(x^4+4 x+1\right)\right] d x; (c) 0π/2[a+(a1)cosθ]1dθ\int_0^{\pi / 2}[a+(a-1) \cos \theta]^{-1} d \theta with a>12a>\frac{1}{2}; (d) (x2+6x+18)1dx\int_{-\infty}^{\infty}\left(x^2+6 x+18\right)^{-1} d x.

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Answered 2 years ago
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(a):0xexdx\textbf{(a)}: \int _{0}^{\infty} x e^{-x} dx

Using by part integration:

=[xex]001ex1dx=[xex]0+0exdx=[xex]0[ex]0=[xex]0[1ex]0=(0e0)(1e1e0)=0+1=1\begin{align*} &=\left[-x e^{-x} \right]_{0}^{\infty} - \int _{0}^{\infty} 1\cdot \dfrac{e^{-x}}{-1} dx\\ &=\left[\dfrac{-x}{e^{x}} \right]_{0}^{\infty}+\int _{0}^{\infty} e^{-x}dx\\\\ &=\left[\dfrac{-x}{e^{x}} \right]_{0}^{\infty}-\left[e^{-x} \right]_{0}^{\infty}\\\\ &=\left[\dfrac{-x}{e^{x}} \right]_{0}^{\infty}- \left[\dfrac{1}{e^{x}} \right]_{0}^{\infty}\\\\ &=\left(\dfrac{-\infty}{\infty}-\dfrac{0}{e^{0}} \right)- \left(\dfrac{1}{e^{\infty}}-\dfrac{1}{e^{0}} \right)\\\\ &= 0+1\\\\ &=\boxed{1} \end{align*}

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