# Find the value of the given integral

\[\int {{a^x}{e^x}dx} \]

Last updated date: 15th Mar 2023

•

Total views: 307.8k

•

Views today: 2.87k

Answer

Verified

307.8k+ views

Hint:- Use the integral by-parts.

Let the value of the given integral be I.

Then, I = \[\int {{a^x}{e^x}dx} \]. (1)

As, we know that if u and v are two functions of $x$ , then the integral of the product of

these two functions will be:

\[ \Rightarrow \int {uvdx = u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\int {vdx} } \right]dx} } } \] (2)

In applying the above equation, the selection of the first function (u) and

Second function (v) should be done depending on which function can be integrated easily.

Normally, we use the preference order for the first function i.e.

ILATE RULE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) which states that the

Inverse function should be assumed as the first function while performing the integration.

Hence the functions are assumed from left to right depending on the type of functions involved.

Then by using the ILATE Rule. We can easily solve the above problem.

According to ILATE Rule,

\[ \Rightarrow u = {a^x}\]

\[ \Rightarrow v = {e^x}\]

So, now putting value of u and v in equation 2 we get,

\[ \Rightarrow I = \int {{a^x}{e^x}dx = {a^x}\int {{e^x}dx - \int {\left[ {\dfrac{{d\left( {{a^x}} \right)}}{{dx}}\int {{e^x}dx} } \right]dx} } } \] (3)

As, we know that, \[\int {{e^x}dx = {e^x}} \]and \[\dfrac{{d\left( {{a^x}} \right)}}{{dx}} = {a^x}.\ln a\]

So, now solving equation 3 we get,

\[ \Rightarrow I = {a^x}.{e^x} - \ln a\int {{a^x}.{e^x}dx} \]

Now, putting the value of \[\int {{a^x}{e^x}dx} \] from equation 1 to above equation. We get,

\[ \Rightarrow I = {a^x}.{e^x} - \ln a(I)\]

Solving above equation we get,

\[

\Rightarrow I\left( {1 + \ln a} \right) = {a^x}.{e^x} \\

\Rightarrow I = \dfrac{{{a^x}.{e^x}}}{{\left( {1 + \ln a} \right)}} \\

\]

Hence the value of given integral is \[\int {{a^x}{e^x}dx} = \dfrac{{{a^x}.{e^x}}}{{\left( {1 + \ln a} \right)}}\].

NOTE:- Whenever we came up with this type of problem then easiest and efficient way to

Solving the problem is using by-parts. And for the selection of the first function we can use ILATE

RULE.Then we can find the value of the given integral using parts. But remember the basic

differentiation and integration formulas.

Let the value of the given integral be I.

Then, I = \[\int {{a^x}{e^x}dx} \]. (1)

As, we know that if u and v are two functions of $x$ , then the integral of the product of

these two functions will be:

\[ \Rightarrow \int {uvdx = u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\int {vdx} } \right]dx} } } \] (2)

In applying the above equation, the selection of the first function (u) and

Second function (v) should be done depending on which function can be integrated easily.

Normally, we use the preference order for the first function i.e.

ILATE RULE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) which states that the

Inverse function should be assumed as the first function while performing the integration.

Hence the functions are assumed from left to right depending on the type of functions involved.

Then by using the ILATE Rule. We can easily solve the above problem.

According to ILATE Rule,

\[ \Rightarrow u = {a^x}\]

\[ \Rightarrow v = {e^x}\]

So, now putting value of u and v in equation 2 we get,

\[ \Rightarrow I = \int {{a^x}{e^x}dx = {a^x}\int {{e^x}dx - \int {\left[ {\dfrac{{d\left( {{a^x}} \right)}}{{dx}}\int {{e^x}dx} } \right]dx} } } \] (3)

As, we know that, \[\int {{e^x}dx = {e^x}} \]and \[\dfrac{{d\left( {{a^x}} \right)}}{{dx}} = {a^x}.\ln a\]

So, now solving equation 3 we get,

\[ \Rightarrow I = {a^x}.{e^x} - \ln a\int {{a^x}.{e^x}dx} \]

Now, putting the value of \[\int {{a^x}{e^x}dx} \] from equation 1 to above equation. We get,

\[ \Rightarrow I = {a^x}.{e^x} - \ln a(I)\]

Solving above equation we get,

\[

\Rightarrow I\left( {1 + \ln a} \right) = {a^x}.{e^x} \\

\Rightarrow I = \dfrac{{{a^x}.{e^x}}}{{\left( {1 + \ln a} \right)}} \\

\]

Hence the value of given integral is \[\int {{a^x}{e^x}dx} = \dfrac{{{a^x}.{e^x}}}{{\left( {1 + \ln a} \right)}}\].

NOTE:- Whenever we came up with this type of problem then easiest and efficient way to

Solving the problem is using by-parts. And for the selection of the first function we can use ILATE

RULE.Then we can find the value of the given integral using parts. But remember the basic

differentiation and integration formulas.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Alfred Wallace worked in A Galapagos Island B Australian class 12 biology CBSE

Imagine an atom made up of a proton and a hypothetical class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main