Question

The value of \[\int {{x^3}{e^x}} dx = \]
A. \[{e^x}\left( {{x^3} - 3{x^2} + 6x + 6} \right) + c\]
B. \[{e^x}\left( {{x^3} - 3{x^2} + 6x - 6} \right) + c\]
C. \[{e^x}\left( {{x^3} + 3{x^2} + 6x + 6} \right) + c\]
D. \[{e^x}\left( {{x^3} - 3{x^2} - 6x + 6} \right) + c\]

Answer 1

Hint: We see that given function is a product of two functions and so To find the value of \[\int {{x^3}{e^x}} dx\], we will use By Parts formula. According to the By Parts formula, \[\int {f(x).g(x)} dx = f(x).\left( {\int {g(x)} dx} \right) - \int {\left( {f'(x).\left( {\int {g(x)} dx} \right)} \right)dx} \]. Here, \[f'(x) = \dfrac{d}{{dx}}f(x)\] and \[f(x)\] is the first function and \[g(x)\] is the second function. First function and second function are decided according to the \[ILATE\] rule. We need to identify the function that comes first on the following list and select it as \[f(x)\]. We need to identify the function that comes first on the following list and select it as \[f(x)\]. \[ILATE\] stands for
\[I - \]Inverse Trigonometric Function
\[L - \]Logarithmic Function
\[A - \]Algebraic Function
\[T - \]Trigonometric Function
\[E - \]Exponential Function

Complete step by step answer:
We need to calculate \[\int {{x^3}{e^x}} dx\] using By Parts formula and \[ILATE\] rule, we see \[{x^3}\] is an algebraic function and \[{e^x}\] is an exponential function. Now, according to \[ILATE\] rule, \[{x^3}\]is our first function and \[{e^x}\] is our second function. Hence, we have
\[f(x) = {x^3} - - - - - - (1)\]
\[\Rightarrow g(x) = {e^x} - - - - - - (2)\]
Using (1) and (2) in the By Parts Formula i.e. \[\int {f(x).g(x)} dx = f(x).\left( {\int {g(x)} dx} \right) - \int {\left( {f'(x).\left( {\int {g(x)} dx} \right)} \right)dx} \], we have
\[\Rightarrow \int {{x^3}.{e^x}} dx = {x^3}.\left( {\int {{e^x}} dx} \right) - \int {\left( {\dfrac{d}{{dx}}\left( {{x^3}} \right).\left( {\int {{e^x}} dx} \right)} \right)dx} \]

We know, \[\int {{e^x}dx = {e^x} + c} \] and \[\dfrac{d}{{dx}}{x^n} = n{\left( x \right)^{n - 1}}\]. Hence, using this
\[ \Rightarrow \int {{x^3}.{e^x}} dx = {x^3}.\left( {{e^x}} \right) - \int {\left( {\left( {3{x^{3 - 1}}} \right).\left( {{e^x}} \right)} \right)dx} \]
\[ \Rightarrow \int {{x^3}.{e^x}} dx = {x^3}.\left( {{e^x}} \right) - \int {\left( {3{x^2}} \right).\left( {{e^x}} \right)dx} \]
We know, \[\int {c.f(x)} dx = c\int {f(x)} dx\], where \[c\] is a constant term.
So, using this in above equation, we get
\[ \Rightarrow \int {{x^3}.{e^x}} dx = {x^3}.\left( {{e^x}} \right) - 3\int {{x^2}{e^x}} dx - - - - - - (3)\]
Now, we see that \[\int {{x^2}{e^x}} dx\]is again a product of two functions and so to solve this, we need to apply the By Parts Formula again.
Now, Solving \[\int {{x^2}{e^x}} dx\]
We see \[{x^2}\] is an algebraic function and \[{e^x}\] is an exponential function. Hence, we take
\[{x^2} = f(x)\]
\[\Rightarrow {e^x} = g(x)\]

Putting these values in \[\int {f(x).g(x)} dx = f(x).\left( {\int {g(x)} dx} \right) - \int {\left( {f'(x).\left( {\int {g(x)} dx} \right)} \right)dx} \], we get
\[\int {{x^2}.{e^x}} dx = {x^2}.\left( {\int {{e^x}} dx} \right) - \int {\left( {\dfrac{d}{{dx}}\left( {{x^2}} \right).\left( {\int {{e^x}} dx} \right)} \right)dx} \]
Using formulas \[\int {{e^x}dx = {e^x} + c} \] and \[\dfrac{d}{{dx}}{x^n} = n{\left( x \right)^{n - 1}}\], we get
\[ \Rightarrow \int {{x^2}.{e^x}} dx = {x^2}.\left( {{e^x}} \right) - \int {\left( {\left( {2{x^{2 - 1}}} \right).\left( {{e^x}} \right)} \right)dx} \]
\[ \Rightarrow \int {{x^2}.{e^x}} dx = {x^2}.\left( {{e^x}} \right) - \int {\left( {\left( {2x} \right).\left( {{e^x}} \right)} \right)dx} \]

Now, using \[\int {c.f(x)} dx = c\int {f(x)} dx\], we have
\[ \Rightarrow \int {{x^2}.{e^x}} dx = {x^2}.\left( {{e^x}} \right) - 2\int {x{e^x}dx} - - - - - - (4)\]
Using (4) in (3), we get
\[\int {{x^3}.{e^x}} dx = {x^3}.\left( {{e^x}} \right) - 3\left( {{x^2}.\left( {{e^x}} \right) - 2\int {x{e^x}dx} } \right)\]
\[ \Rightarrow \int {{x^3}.{e^x}} dx = {x^3}{e^x} - 3{x^2}{e^x} + 6\int {x{e^x}dx} - - - - - - (5)\]
Again, we see that \[\int {x{e^x}dx} \] is the product of two functions and so to find \[\int {x{e^x}dx} \], we need to use By Parts Formula again. In \[x{e^x}\], we see that \[x\] is an algebraic function and \[{e^x}\] is an exponential function. Hence, according to \[ILATE\] rule, we take
\[f(x) = x\]
\[\Rightarrow g(x) = {e^x}\]

Using these values in \[\int {f(x).g(x)} dx = f(x).\left( {\int {g(x)} dx} \right) - \int {\left( {f'(x).\left( {\int {g(x)} dx} \right)} \right)dx} \], we get
\[\int {x.{e^x}} dx = x.\left( {\int {{e^x}} dx} \right) - \int {\left( {\dfrac{d}{{dx}}\left( x \right).\left( {\int {{e^x}} dx} \right)} \right)dx} \]
Using formulas \[\int {{e^x}dx = {e^x} + c} \] and \[\dfrac{d}{{dx}}{x^n} = n{\left( x \right)^{n - 1}}\], we get
\[ \Rightarrow \int {x.{e^x}} dx = x.\left( {{e^x}} \right) - \int {\left( {\left( {1{x^{1 - 1}}} \right).\left( {{e^x}} \right)} \right)dx} \]
\[ \Rightarrow \int {x.{e^x}} dx = x.\left( {{e^x}} \right) - \int {\left( {\left( {1{x^0}} \right).\left( {{e^x}} \right)} \right)dx} \]

We know, \[{x^0} = 1\]. So, the equation becomes
\[ \Rightarrow \int {x.{e^x}} dx = x.\left( {{e^x}} \right) - \int {\left( {1.{e^x}} \right)dx} \]
\[ \Rightarrow \int {x.{e^x}} dx = x.\left( {{e^x}} \right) - \int {{e^x}dx} \]
Again using \[\int {{e^x}dx = {e^x} + c} \], we get
\[ \Rightarrow \int {x.{e^x}} dx = x.\left( {{e^x}} \right) - {e^x} + c - - - - - - (6)\]
Using (6) in (5), we get
\[\int {{x^3}.{e^x}} dx = {x^3}{e^x} - 3{x^2}{e^x} + 6\left( {x.\left( {{e^x}} \right) - {e^x}} \right) + c\], where \[c\] is a constant term.
\[ \Rightarrow \int {{x^3}.{e^x}} dx = {x^3}{e^x} - 3{x^2}{e^x} + 6{e^x}x - 6{e^x} + c\], where \[c\] is a constant term.
Now taking out \[{e^x}\]common from first four terms, we get
\[ \therefore \int {{x^3}.{e^x}} dx = {e^x}\left( {{x^3} - 3{x^2} + 6x - 6} \right) + c\]

Therefore, the correct answer is B.

Note: In this we need to use the By Parts formula again and again. We need to be careful while selecting the first function and the second function. As wrong function selection will give us the wrong answer. Also, while using the By Parts formula, we should take care of the second term i.e. we need to first integrate the second function and then integrate the product of derivatives of the first function and integration of the second function. While combining the terms and putting the values of one term in another one, we should be very careful with the signs.