
The value of the integral $\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} $is
$
{\text{a}}{\text{. }}{{\text{e}}^x}f'\left( x \right) + c \\
{\text{b}}{\text{. }}{{\text{e}}^x}f\left( x \right) + c{\text{ }} \\
{\text{c}}{\text{. }}\dfrac{{{e^x}}}{{f\left( x \right)}} + c \\
{\text{d}}{\text{. }}\dfrac{{{e^x}}}{{f'\left( x \right)}} + c \\
$
Answer
634.5k+ views
Hint: - Apply integration by parts method to reach the answer which is given as
$\int {\left( {{I_1}} \right)\left( {{I_2}} \right)dx = {I_1}} \int {\left( {{I_2}} \right)dx - \int {\left( {\left( {\dfrac{d}{{dx}}{I_1}} \right)\int {\left( {{I_2}} \right)dx} } \right)} } dx$
Given integral is
$I = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} $
Now break the integration into two parts
$ \Rightarrow I = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx + \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} } $
Let $I = {I_1} + {I_2}................................\left( 1 \right)$
First solve ${I_1}$
${I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} $
Now integrate it by parts where $f\left( x \right)$ is the first part and ${e^x}$ is the second part.
$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = f\left( x \right)\int {{e^x}dx - \int {\left( {\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)\left( {\int {{e^x}dx} } \right)} \right)dx} } $
As we know$\int {{e^x}dx = {e^x},{\text{ }}} \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)$, so apply this property
$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c} $, (where c is some arbitrary integration constant)
From equation (1)
$
I = {I_1} + {I_2} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c + } \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} \\
\Rightarrow I = {e^x}f\left( x \right) + c \\
$
Hence, option (b) is correct.
Note: - In such types of questions always remember the key rule that first breaks the integration into parts, then integrate first integration by integration by parts method and leave the second integration as it is, after that it cancels out the second integration, doing this we will get the required answer.
$\int {\left( {{I_1}} \right)\left( {{I_2}} \right)dx = {I_1}} \int {\left( {{I_2}} \right)dx - \int {\left( {\left( {\dfrac{d}{{dx}}{I_1}} \right)\int {\left( {{I_2}} \right)dx} } \right)} } dx$
Given integral is
$I = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} $
Now break the integration into two parts
$ \Rightarrow I = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx + \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} } $
Let $I = {I_1} + {I_2}................................\left( 1 \right)$
First solve ${I_1}$
${I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} $
Now integrate it by parts where $f\left( x \right)$ is the first part and ${e^x}$ is the second part.
$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = f\left( x \right)\int {{e^x}dx - \int {\left( {\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)\left( {\int {{e^x}dx} } \right)} \right)dx} } $
As we know$\int {{e^x}dx = {e^x},{\text{ }}} \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)$, so apply this property
$ \Rightarrow {I_1} = \int {{e^x}\left\{ {f\left( x \right)} \right\}dx} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c} $, (where c is some arbitrary integration constant)
From equation (1)
$
I = {I_1} + {I_2} = {e^x}f\left( x \right) - \int {{e^x}f'\left( x \right)dx + c + } \int {{e^x}\left\{ {f'\left( x \right)} \right\}dx} \\
\Rightarrow I = {e^x}f\left( x \right) + c \\
$
Hence, option (b) is correct.
Note: - In such types of questions always remember the key rule that first breaks the integration into parts, then integrate first integration by integration by parts method and leave the second integration as it is, after that it cancels out the second integration, doing this we will get the required answer.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

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

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

Coming together federation is practiced in A India class 12 social science CBSE

Write the formula to find the shortest distance between class 12 maths CBSE

