Question

Integrate \[\int {x{e^{2x}}} dx\]

Answer 1

Hint: Integration is the calculation of an integral. These are used in maths to find the magnitude of certain physical quantities, example area, volume, displacement. Since it is the opposite of differentiation therefore, it is called antiderivative.
Formula used: \[\int {u\,v\,dx = u\int {v\,dx - } } \int {u`(\int {v\,dx)\,dx} } \]

Complete step-by-step solution:
We solve this question by integration by-parts, this way of integration is often useful when two functions are multiplied together, but is also helpful in other ways
By using integration by-parts we get,
\[u = x \Rightarrow \,du = dx\,\]
And, \[dv = {e^{2x}}dx \Rightarrow v = \dfrac{{{e^{2x}}}}{2}\]
Now, when we solve this equation further, we get:
\[\int {u.vdx = u\int {vdx - \int {\left[ {\int {vdx.\dfrac{{du}}{{dx}}.dx} } \right]} } } \]
Now, we will apply the by-parts formula, and we get:
\[ \Rightarrow I = \dfrac{{x{e^{2x}}}}{2} - \int {\dfrac{{{e^{2x}}}}{2}} \times {e^{2x}}dx\]
\[ \Rightarrow I = \dfrac{{x{e^{2x}}}}{2} - \dfrac{1}{2}\int {{e^{2x}}} dx\]
Now we will try to solve it further. By further calculating, we get:
\[\Rightarrow I = \dfrac{{x{e^{2x}}}}{2} - \dfrac{1}{2} \times \dfrac{{{e^{2x}}}}{2} + c \]
Therefore, we get that:
  \[I = \dfrac{{x{e^{2x}}}}{2} - \dfrac{{{e^{2x}}}}{4} + c \]
So, our result is that:
\[\int {x{e^{2x}}} dx = \dfrac{{x{e^{2x}}}}{2} - \dfrac{{{e^{2x}}}}{4} + c\]

Additional information: Integration is actually the opposite of differentiation. There are two types of integral. One is a definite integral in which the integral contains lower and upper limits. The second one is the indefinite integral in which it does not have any limits.

Note: There are two cases where integration by-parts is applied. One is the logarithmic function in ‘x’. The second was the first four inverse trigonometric functions (arcsin x, across x, arctan x, arccos x). While solving the question, signs and values should be kept properly and look if there are any limits that are to be applied.