Math, asked by benifa, 5 months ago

evaluate integral x²e^xdx​

Answers

Answered by Asterinn
13

 \rm \longrightarrow \displaystyle \int \rm {x}^{2}  {e}^{x}  \: dx \\  \\  \\ \rm \longrightarrow {x}^{2}\displaystyle \int \rm   {e}^{x}  \: dx - \int \rm \bigg(  \frac{d( {x}^{2} )}{dx}   \int{e}^{x}  \: dx \bigg) dx\\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - \int \rm  2x   \: {e}^{x}  dx\\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2\int \rm  x   \: {e}^{x}  dx

\rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2\int \rm  x   \: {e}^{x}  dx \\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2 \bigg[x\int \rm    \: {e}^{x}  dx  -  \int \bigg( \frac{dx}{dx}  \int {e}^{x}  dx \bigg)dx\bigg]\\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2 \bigg[x \rm    \: {e}^{x}  -  \int {e}^{x} dx\bigg]\\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2 (x \rm    \: {e}^{x}  -   {e}^{x})+ c\\  \\  \\ \rm \longrightarrow {x}^{2}\rm   {e}^{x}   - 2 x \rm    \: {e}^{x}  -   2{e}^{x}+ c


Anonymous: Nice ☃️
Anonymous: Nicee as always ❤:)
Answered by mathdude500
5

\large\underline\blue{\bold{Given \:  Question :-  }}

\bf \:Evaluate : \bf\int\limits {x}^{2}  {e}^{x} dx

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\large\underline\blue{\bold{Formula \:  used:-  }}

✏️Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

✏️See the rule:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

  • 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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\huge {AηsωeR} ✍

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\bf\int\limits {x}^{2}  {e}^{x} dx

✏️Here,

\sf \:  ⟼u =  {x}^{2}  \: and \: v \:  =  {e}^{x}

\underline{\boxed{\star{\sf{\blue{ Using  \: Integration  \: by  \: Parts  \: we \: get}}}}}

\sf \:   =  {x}^{2} \sf\int\limits{e}^{x} dx - \bf\int\limits(\dfrac{d}{dx}  {x}^{2} \sf\int\limits{e}^{x} dx)dx

\sf \:   =  {x}^{2} {e}^{x}  - \sf\int\limits2x{e}^{x} dx

\sf \:   =  {x}^{2} {e}^{x}  - 2\sf\int\limits \: x{e}^{x} dx

\underline{\boxed{\star{\sf{\blue{ Using  \: Integration  \: by  \: Parts  \: in \: second \: term \: we \: get}}}}}

\sf \:   =  {x}^{2} {e}^{x}  -2(x\sf\int\limits{e}^{x} dx - \sf\int\limits(\dfrac{d}{dx} x\sf\int\limits{e}^{x} dx)dx)

\sf \:   =  {x}^{2} {e}^{x}  -2x{e}^{x}  + 2\sf\int\limits{e}^{x} dx

\sf \:   =  {x}^{2} {e}^{x}  -2x{e}^{x}  +2{e}^{x}  + c

\underline{\boxed{\star{\sf{\blue{ Hence, \bf\int\limits {x}^{2}  {e}^{x} dx = \sf \:   =  {x}^{2} {e}^{x}  -2x{e}^{x}  +2{e}^{x}  + c}}}}}

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