Question

If\[f\left( x \right) = \int_a^x {{t^3}{e^t}dt} \], then find the value of\[\dfrac{d}{{dx}}f\left( x \right)\].
A. \[{e^x}\left( {{x^3} + 3{x^2}} \right)\]
B. \[{x^3}{e^x}\]
C. \[{a^3}{e^a}\]
D. None of these

Answer 1

Hint: In this question, we are given an integral as\[f\left( x \right) = \int_a^x {{t^3}{e^t}dt} \]. First, we will evaluate the given integral by using the identity\[\int_a^x {f\left( x \right)dx} = \int_a^0 {f\left( x \right)dx} + \int_0^x {f\left( x \right)dx} \].
After that, we will integrate by parts \[\int {fg' = fg - } \int {f'g} \]
Now, we will integrate from -1 to 0 and then from 0 to 1. At last, we will put the limits value to get the final result.
Finally we will find the derivative of resultant value of the function\[f\left( x \right)\]with respect to\[x\].



Formula Used:We will use the following formulas to solve the given problems:
1) \[\int_a^b {f\left( x \right)dx = } \int_a^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} ,\left( {a < c < b} \right)\]
2) \[\int {fg' = fg - } \int {f'g} \]



Complete step by step solution:We have been given\[f\left( x \right) = \int_a^x {{t^3}{e^t}dt} \].
First, we will evaluate the given integral by using the identity\[\int_a^x {f\left( x \right)dx} = \int_a^0 {f\left( x \right)dx} + \int_0^x {f\left( x \right)dx} \].
\[f\left( x \right) = \int_a^0 {{t^3}{e^t}dt} + \int_0^x {{t^3}{e^t}dt} \]
\[ \Rightarrow f\left( x \right) = {I_1} + {I_2}\] ……… (1)
We will integrate\[{I_1}\]and \[{I_2}\]by parts: \[\int {fg' = fg - } \int {f'g} \]
For\[{I_1}\]:\[f = {t^3}\],\[f' = 3{t^2}\], \[g = {e^t}\],\[g' = {e^t}\]
\[{I_1} = {t^3}{e^t} - \int {3{t^2}{e^t}dt} \]
Again integrate \[\int {3{t^2}{e^t}dt} \]by parts.
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3\left[ {\int {{t^2}{e^t}dt} } \right]\]
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3\left[ {{t^2}{e^t} - \int {2t{e^t}dt} } \right]\]
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3\left[ {{t^2}{e^t} - 2\int {t{e^t}dt} } \right]\]
We will again integrate this\[\int {t{e^t}dt} \]by parts.
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3\left[ {{t^2}{e^t} - 2\left( {t{e^t} - \int {{e^t}dt} } \right)} \right]\]
As we know that\[\int {{e^t}dt} = \dfrac{{{e^t}}}{{\ln \left( e \right)}} = {e^t}\], then we get
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3\left[ {{t^2}{e^t} - \left( {2t{e^t} - 2{e^t}} \right)} \right]\]
\[ \Rightarrow {I_1} = {t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t} + C\]
\[ \Rightarrow {I_1} = \left( {{t^3} - 3{t^2} + 6t - 6} \right){e^t} + {C_1}\]
When we integrate from \[a\] to\[0\], we get
\[ \Rightarrow {I_1} = - \left( {{a^3} - 3{a^2} + 6a - 6} \right){e^a} + {C_1}\]
Similarly, we will integrate \[{I_2}\]by parts: \[\int {fg' = fg - } \int {f'g} \]
\[ \Rightarrow {I_2} = \left( {{t^3} - 3{t^2} + 6t - 6} \right){e^t} + {C_2}\]
When we integrate from \[0\] to\[x\], we get
\[ \Rightarrow {I_2} = \left( {{x^3} - 3{x^2} + 6x - 6} \right){e^x} + {C_2}\]
We will further substitute the values of \[{I_1}\]and \[{I_2}\]in equation 1 to get
\[ \Rightarrow f\left( x \right) = - \left( {{a^3} - 3{a^2} + 6a - 6} \right){e^a} + {C_1} + \left( {{x^3} - 3{x^2} + 6x - 6} \right){e^x} + {C_2}\]
As we can write\[{C_1} + {C_2} = C\]
\[ \Rightarrow f\left( x \right) = - \left( {{a^3} - 3{a^2} + 6a - 6} \right){e^a} + \left( {{x^3} - 3{x^2} + 6x - 6} \right){e^x} + C\]
Finally we will find the derivative of function\[f\left( x \right)\]with respect to\[x\].
\[\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ { - \left( {{a^3} - 3{a^2} + 6a - 6} \right){e^a} + \left( {{x^3} - 3{x^2} + 6x - 6} \right){e^x} + C} \right]\]
By using chain rule\[\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{d}{{dx}}\left( v \right) + v\dfrac{d}{{dx}}\left( u \right)\], we get
\[\begin{array}{l}\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \left( { - {a^3} + 3{a^2} - 6a + 6} \right)\dfrac{d}{{dx}}\left( {{e^a}} \right) + {e^a}\dfrac{d}{{dx}}\left( { - {a^3} + 3{a^2} - 6a + 6} \right) + \left( {{x^3} - 3{x^2} + 6x - 6} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)\\ + {e^x}\dfrac{d}{{dx}}\left( {{x^3} - 3{x^2} + 6x - 6} \right)\end{array}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \left( { - {a^3} + 3{a^2} - 6a + 6} \right)\left( 0 \right) + {e^a}\left( 0 \right) + \left( {{x^3} - 3{x^2} + 6x - 6} \right){e^x} + {e^x}\left( {3{x^2} - 6x + 6} \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = {x^3}{e^x} - 3{x^2}{e^x} + 6x{e^x} - 6{e^x} + 3{x^2}{e^x} - 6x{e^x} + 6{e^x}\]
After further simplification, we get
\[\dfrac{d}{{dx}}f\left( x \right) = {x^3}{e^x}\]
As a result, the value of\[\dfrac{d}{{dx}}f\left( x \right)\]is\[{x^3}{e^x}\].



Option ‘B’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly \[x\] to\[a\]. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier.