Question

Evaluate this integral

$\int \limits _0^ \infty t {e}^{ - t} dt$

​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \int^{ \infty } _{0} \: t {e}^{ - t} \: dt$

Solving by IBP,

$\displaystyle = \left[ t\int {e}^{ - t} \: dt \right]^{ \infty } _{0} - \int^{ \infty } _{0} \left \{ \dfrac{d}{dt}(t) \cdot\int {e}^{ - t} \: dt\right \} dt$

$\displaystyle = - \left[ t \cdot {e}^{ - t} \right]^{ \infty } _{0} - \int^{ \infty } _{0} \left \{ 1 \cdot( - {e}^{ - t} )\right \} dt$

$\displaystyle = - \left[ t \cdot {e}^{ - t} \right]^{ \infty } _{0} + \int^{ \infty } _{0} {e}^{ - t} dt$

$\displaystyle = - \left[ t \cdot {e}^{ - t} \right]^{ \infty } _{0} - [ {e}^{ - t} ]^{ \infty } _{0}$

$\displaystyle = - 0 - [ {e}^{ - \infty } - {e}^{0} ]$

$\displaystyle = - [ 0 - 1 ] = 1$

Answer 2

$\green{\large\underline{\sf{Solution-}}}$

Given integral is

$\red{\rm :\longmapsto\: \displaystyle\int \limits _0^ \infty t {e}^{ - t} dt}$

Let first evaluate

$\rm :\longmapsto\:\displaystyle\int t {e}^{ - t} \: dt$

We know, integration by parts

$\displaystyle\int uvdx \: = u\displaystyle\int vdx - \displaystyle\int \bigg[\dfrac{d}{dx}u\displaystyle\int vdx \bigg]dx$

So, using this, we get

$\rm \:  =  \: t\displaystyle\int {e}^{ - t}dt - \displaystyle\int \bigg[\dfrac{d}{dt}t\displaystyle\int {e}^{ - t}dt\bigg]dt$

$\rm \:  =  \: - \: t {e}^{ - t} + \displaystyle\int {e}^{ - t}dt$

$\rm \:  =  \: - \: t {e}^{ - t} - {e}^{ - t}$

$\rm \:  =  \: - \: [t - 1]{e}^{ - t}$

So, we get

$\rm :\longmapsto\:\displaystyle\int t {e}^{ - t} \: dt = [t - 1]{e}^{ - t}$

Hence,

$\rm :\longmapsto\:\displaystyle\int \limits _0^ \infty t {e}^{ - t} dt = \bigg[(t - 1){e}^{ - t} \bigg]_0^ \infty$

$\rm \:  =  \: {e}^{ - \infty } - (0 - 1) {e}^{0}$

$\rm \:  =  \: 0 + 1$

$\rm \:  =  \: 1$

Hence,

$\red{\rm :\longmapsto\: \boxed{\tt{ \: \: \: \: \: \: \: \: \: \displaystyle\int \limits _0^ \infty t {e}^{ - t} dt = 1 \: \: \: \: \: \: \: \: }}}$

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Basic Concept Used

Integration by Parts

$\displaystyle\int uvdx \: = u\displaystyle\int vdx - \displaystyle\int \bigg[\dfrac{d}{dx}u\displaystyle\int vdx \bigg]dx$

Where,

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

• I - Inverse trigonometric functions

• L -Logarithmic functions

• A - Arithmetic and Algebraic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is choosen as u and other as v.

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$