
Evaluate the following integral:
\[\int {x{e^{{x^2}}}dx} \]
Answer
520.5k+ views
Hint: This problem can be solved by a substitution method. The substitution method (also called u-substitution) is used when an integral contains some function and its derivative. In this case, we can set u equal to the function and rewrite the integral in terms of new variable u. This makes integral easier to solve. So, use this concept to reach the solution of the problem.
Complete step-by-step answer:
Let \[I = \int {x{e^{{x^2}}}dx} \]
Put \[{x^2} = t\] and differentiate it w.r.t ‘\[x\]’
\[
\Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = \dfrac{{d\left( t \right)}}{{dx}} \\
\Rightarrow 2x = \dfrac{{dt}}{{dx}} \\
\Rightarrow 2xdx = dt \\
\therefore xdx = \dfrac{{dt}}{2} \\
\]
Then \[I = \int {{e^t}\dfrac{{dt}}{2}} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{e^t}dt} \]
We know that \[\int {{e^x}dx} = {e^x} + c\]
\[
\Rightarrow I = \dfrac{1}{2}\left( {{e^t} + c} \right) \\
\Rightarrow I = \dfrac{1}{2}{e^t} + \dfrac{c}{2} \\
\therefore I = \dfrac{1}{2}{e^t} + C \\
\]
Since, \[{x^2} = t\]
Therefore, \[I = \dfrac{1}{2}{e^{{x^2}}} + C\]
Thus, \[\int {x{e^{{x^2}}}dx} = \dfrac{1}{2}{e^{{x^2}}} + C\]
Note: A constant namely integrating constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. So, it is necessary to add integrating constants after completion of the integral.
Complete step-by-step answer:
Let \[I = \int {x{e^{{x^2}}}dx} \]
Put \[{x^2} = t\] and differentiate it w.r.t ‘\[x\]’
\[
\Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = \dfrac{{d\left( t \right)}}{{dx}} \\
\Rightarrow 2x = \dfrac{{dt}}{{dx}} \\
\Rightarrow 2xdx = dt \\
\therefore xdx = \dfrac{{dt}}{2} \\
\]
Then \[I = \int {{e^t}\dfrac{{dt}}{2}} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{e^t}dt} \]
We know that \[\int {{e^x}dx} = {e^x} + c\]
\[
\Rightarrow I = \dfrac{1}{2}\left( {{e^t} + c} \right) \\
\Rightarrow I = \dfrac{1}{2}{e^t} + \dfrac{c}{2} \\
\therefore I = \dfrac{1}{2}{e^t} + C \\
\]
Since, \[{x^2} = t\]
Therefore, \[I = \dfrac{1}{2}{e^{{x^2}}} + C\]
Thus, \[\int {x{e^{{x^2}}}dx} = \dfrac{1}{2}{e^{{x^2}}} + C\]
Note: A constant namely integrating constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. So, it is necessary to add integrating constants after completion of the integral.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
