Solve the given indefinite integral:\[\int {{x^4}{e^{2x}}dx = } \]
A. \[\dfrac{{{e^{2x}}}}{4}\left( {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right) + c\]
B. \[\dfrac{{{e^{2x}}}}{2}\left( {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right) + c\]
C. \[\dfrac{{{e^{2x}}}}{8}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + c\]
D. \[\dfrac{{{e^{2x}}}}{4}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + c\]
Answer
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Hint: To solve the given question above we will first put \[{\text{2x}} = {\text{t}}\] and convert whole integrals in terms of t. Now use the integration by parts method to integrate the given function and at last replace t with its original value.
Complete step-by-step answer:
In the question given, we have to find the value of \[\int {{e^{2x}}.{x^4}dx} \]. Let its value be I. Thus, we have:
\[I = \int {{x^4}{e^{2x}}dx} \]
Now, we will put \[2x = t\] .On differentiating both sides we will get:
\[2dx = dt\]
\[ \Rightarrow dx = \dfrac{{dt}}{2}\]
Thus, we will get:
\[\begin{array}{l}I = \int {{{\left( {\dfrac{t}{2}} \right)}^4}{e^t}\left( {\dfrac{{dt}}{2}} \right)} \\I = \left( {\dfrac{1}{{{2^5}}}} \right)\int {{e^t}.{t^4}dt} \end{array}\]
Now, we will calculate the value of \[\int {{e^t}.t} \] with the help of by parts. According to by parts theorem of integration, we have:
\[\int {uv = } u\int {v - \int {u\int v } } \]
In our case \[u = t\,\,and\,\,v - {e^t}\] . Thus, we will get:
\[\begin{array}{l}\int {{e^t}.t\,dt = } \,t\int {{e^t}\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {{e^t}\,\left( {dt} \right)} }^2}} } \\\int {{e^t}t\,dt = } \,t{e^t} - \int {{e^t}\,\left( {dt} \right)} \end{array}\]
\[\int {{e^t}t\,dt = } \,t{e^t} - {e^t} + {c_1}\] . . . . . . . . . . . . . (i)
Now we will calculate the value of \[\int {{e^t}{t^t}\,dt} \]. In this function, we will again use by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}t\] . Thus, we will get:
\[\int {{e^t}{t^2}\,dt = } \,t\int {{e^t}t\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {\left( {{e^t}t\,} \right)\left( {dt} \right)} }^2}} } \] . . . . . . . . . . . . . (ii)
Now, we will put the value of \[\int {{e^t}t\,dt} \] from (i) into (ii). Thus, we will get following equation:
\[\begin{array}{l}\int {{e^t}{t^2}\,dt = } \,t\left[ {t{e^t} - {e^t}} \right] - \int {\left( {t{e^t} - {e^t}} \right)} \,dt + {c_2}\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \int {t{e^t}dt + \int {{e^t}} dt + {c_2}} \,\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \left( {t{e^t} - {e^t}} \right)\, + {e^t} + {c_2}\end{array}\]
\[\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - 2t{e^t}\, + 2{e^t} + {c_2}\] . . . . . . . . . . . . . (iii)
Now, we will calculate the value of \[\int {{e^t}{t^3}} \] by the help of by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}{t^2}\] . Thus, we will get:
\[\int {{e^t}{t^3}\,dt = } \,t\int {{e^t}{t^2}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^2}} \right)} } \,{\left( {dt} \right)^2}\] . . . . . . . . . . . . . (iv)
Now, we will put the value of \[{e^t}{t^2}\] from (iii) to (iv). Thus, we will get:
\[\begin{array}{l} \Rightarrow \int {{e^t}{t^3}\,dt = } \,t\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - \int {\left( {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right)} \,dt + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \int {{t^2}{e^t}dt + 2\int {t{e^t}dt} - 2\int {{e^t}dt} \,} + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] + 2\left[ {t{e^t} - {e^t}} \right] - 2{e^t} + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - {t^2}{e^t} + 2t{e^t} - 2{e^t} + 2t{e^t} - 2{e^t} - 2{e^t} + {c_3}\end{array}\]
\[ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 3{t^2}{e^t} - 6t{e^t} - 6{e^t} + {c_3}\] . . . . . . . . . . . . . (v)
Now, we will calculate the value of \[{e^t}{t^4}dt\] by the help of by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}{t^3}\] . Thus, we will get:
\[\int {{e^t}{t^4}\,dt = } \,t\int {{e^t}{t^3}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^3}} \right)} } \,{\left( {dt} \right)^2}\] . . . . . . . . . . . . (vi)
Now, we will put the value of \[{e^t}{t^3}\,dt\] from (v) to (vi). Thus, we will get:
\[\begin{array}{l} \Rightarrow \int {{e^t}{t^4}\,dt = } \,t\left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] - \int {\left( {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right)} \,dt + {c_4}\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \int {{t^3}{e^t}dt + \int {3{t^2}{e^t}dt} - 6\int {t{e^t}dt} \,} + 6\int {{e^t}dt} \, + {c_4}\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] + 3\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - 6\left[ {t{e^t} - {e^t}} \right] + 6{e^t} + {c_4}\end{array}\]\[ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t} + {c_4}\] . . . . . . . . . . . . (vii)
Now, we will put the value of \[\int {{e^t}{t^4}} \] from (vii) to I. Thus, we will get the following equation:
\[\begin{array}{l} \Rightarrow I = \dfrac{1}{{{2^5}}}\,\left[ {{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t}} \right] + c\\ \Rightarrow I = \dfrac{{{e^t}}}{{32}}\,\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}\]
Now, we will put the value of y back to 2x. Thus, we will get the following result:
\[\begin{array}{l} \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {{{\left( {2x} \right)}^4} - 4{{\left( {2x} \right)}^3} + 12{{\left( {2x} \right)}^2} - 24\left( {2x} \right) + 24} \right] + c\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {16{x^4} - 32{x^3} + 48{x^2} - 48x + 24} \right] + c\\ \Rightarrow I = \dfrac{{8{e^{2x}}}}{{32}}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{4}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\end{array}\]
Hence, option (a) is correct.
Note: The shortcut method for finding the integrals of \[{e^t}{t^n}\], where n is an integer is given below:
\[\int {{e^t}{t^n}\, = } \,{e^t}\left[ {{t^n} - \dfrac{d}{{dt}}\left( {{t^n}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^n}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^n}} \right) + ............{{\left( { - 1} \right)}^n}\dfrac{{{d^n}}}{{d{t^n}}}\left( {{t^n}} \right)} \right]\]
In our case \[{\text{n}} = {\text{4}}\] so, we will get:
\[\begin{array}{l}\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - \dfrac{d}{{dt}}\left( {{t^4}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^4}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^4}} \right) + \dfrac{{{d^4}}}{{d{t^4}}}\left( {{t^4}} \right)} \right] + c\\\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}\]
Complete step-by-step answer:
In the question given, we have to find the value of \[\int {{e^{2x}}.{x^4}dx} \]. Let its value be I. Thus, we have:
\[I = \int {{x^4}{e^{2x}}dx} \]
Now, we will put \[2x = t\] .On differentiating both sides we will get:
\[2dx = dt\]
\[ \Rightarrow dx = \dfrac{{dt}}{2}\]
Thus, we will get:
\[\begin{array}{l}I = \int {{{\left( {\dfrac{t}{2}} \right)}^4}{e^t}\left( {\dfrac{{dt}}{2}} \right)} \\I = \left( {\dfrac{1}{{{2^5}}}} \right)\int {{e^t}.{t^4}dt} \end{array}\]
Now, we will calculate the value of \[\int {{e^t}.t} \] with the help of by parts. According to by parts theorem of integration, we have:
\[\int {uv = } u\int {v - \int {u\int v } } \]
In our case \[u = t\,\,and\,\,v - {e^t}\] . Thus, we will get:
\[\begin{array}{l}\int {{e^t}.t\,dt = } \,t\int {{e^t}\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {{e^t}\,\left( {dt} \right)} }^2}} } \\\int {{e^t}t\,dt = } \,t{e^t} - \int {{e^t}\,\left( {dt} \right)} \end{array}\]
\[\int {{e^t}t\,dt = } \,t{e^t} - {e^t} + {c_1}\] . . . . . . . . . . . . . (i)
Now we will calculate the value of \[\int {{e^t}{t^t}\,dt} \]. In this function, we will again use by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}t\] . Thus, we will get:
\[\int {{e^t}{t^2}\,dt = } \,t\int {{e^t}t\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {\left( {{e^t}t\,} \right)\left( {dt} \right)} }^2}} } \] . . . . . . . . . . . . . (ii)
Now, we will put the value of \[\int {{e^t}t\,dt} \] from (i) into (ii). Thus, we will get following equation:
\[\begin{array}{l}\int {{e^t}{t^2}\,dt = } \,t\left[ {t{e^t} - {e^t}} \right] - \int {\left( {t{e^t} - {e^t}} \right)} \,dt + {c_2}\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \int {t{e^t}dt + \int {{e^t}} dt + {c_2}} \,\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \left( {t{e^t} - {e^t}} \right)\, + {e^t} + {c_2}\end{array}\]
\[\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - 2t{e^t}\, + 2{e^t} + {c_2}\] . . . . . . . . . . . . . (iii)
Now, we will calculate the value of \[\int {{e^t}{t^3}} \] by the help of by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}{t^2}\] . Thus, we will get:
\[\int {{e^t}{t^3}\,dt = } \,t\int {{e^t}{t^2}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^2}} \right)} } \,{\left( {dt} \right)^2}\] . . . . . . . . . . . . . (iv)
Now, we will put the value of \[{e^t}{t^2}\] from (iii) to (iv). Thus, we will get:
\[\begin{array}{l} \Rightarrow \int {{e^t}{t^3}\,dt = } \,t\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - \int {\left( {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right)} \,dt + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \int {{t^2}{e^t}dt + 2\int {t{e^t}dt} - 2\int {{e^t}dt} \,} + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] + 2\left[ {t{e^t} - {e^t}} \right] - 2{e^t} + {c_3}\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - {t^2}{e^t} + 2t{e^t} - 2{e^t} + 2t{e^t} - 2{e^t} - 2{e^t} + {c_3}\end{array}\]
\[ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 3{t^2}{e^t} - 6t{e^t} - 6{e^t} + {c_3}\] . . . . . . . . . . . . . (v)
Now, we will calculate the value of \[{e^t}{t^4}dt\] by the help of by-parts. In this case, we will take \[u = t\,\,and\,\,v = {e^t}{t^3}\] . Thus, we will get:
\[\int {{e^t}{t^4}\,dt = } \,t\int {{e^t}{t^3}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^3}} \right)} } \,{\left( {dt} \right)^2}\] . . . . . . . . . . . . (vi)
Now, we will put the value of \[{e^t}{t^3}\,dt\] from (v) to (vi). Thus, we will get:
\[\begin{array}{l} \Rightarrow \int {{e^t}{t^4}\,dt = } \,t\left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] - \int {\left( {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right)} \,dt + {c_4}\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \int {{t^3}{e^t}dt + \int {3{t^2}{e^t}dt} - 6\int {t{e^t}dt} \,} + 6\int {{e^t}dt} \, + {c_4}\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] + 3\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - 6\left[ {t{e^t} - {e^t}} \right] + 6{e^t} + {c_4}\end{array}\]\[ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t} + {c_4}\] . . . . . . . . . . . . (vii)
Now, we will put the value of \[\int {{e^t}{t^4}} \] from (vii) to I. Thus, we will get the following equation:
\[\begin{array}{l} \Rightarrow I = \dfrac{1}{{{2^5}}}\,\left[ {{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t}} \right] + c\\ \Rightarrow I = \dfrac{{{e^t}}}{{32}}\,\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}\]
Now, we will put the value of y back to 2x. Thus, we will get the following result:
\[\begin{array}{l} \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {{{\left( {2x} \right)}^4} - 4{{\left( {2x} \right)}^3} + 12{{\left( {2x} \right)}^2} - 24\left( {2x} \right) + 24} \right] + c\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {16{x^4} - 32{x^3} + 48{x^2} - 48x + 24} \right] + c\\ \Rightarrow I = \dfrac{{8{e^{2x}}}}{{32}}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{4}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\end{array}\]
Hence, option (a) is correct.
Note: The shortcut method for finding the integrals of \[{e^t}{t^n}\], where n is an integer is given below:
\[\int {{e^t}{t^n}\, = } \,{e^t}\left[ {{t^n} - \dfrac{d}{{dt}}\left( {{t^n}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^n}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^n}} \right) + ............{{\left( { - 1} \right)}^n}\dfrac{{{d^n}}}{{d{t^n}}}\left( {{t^n}} \right)} \right]\]
In our case \[{\text{n}} = {\text{4}}\] so, we will get:
\[\begin{array}{l}\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - \dfrac{d}{{dt}}\left( {{t^4}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^4}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^4}} \right) + \dfrac{{{d^4}}}{{d{t^4}}}\left( {{t^4}} \right)} \right] + c\\\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}\]
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