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Question

Question: Evaluate the integral \[\int {x{e^{ - a{x^2}}}} dx\]?...

Evaluate the integral xeax2dx\int {x{e^{ - a{x^2}}}} dx?

Explanation

Solution

Here we have to find the integral xeax2dx\int {x{e^{ - a{x^2}}}} dx. For this, we will substitute u=ax2u = - a{x^2}. Then we will differentiate u=ax2u = - a{x^2} and substitute xdx=du2axdx = - \dfrac{{du}}{{2a}}. Then as we know that exdx=ex+C\int {{e^x}} dx = {e^x} + C, using this we will integrate and at last we will substitute back u=ax2u = - a{x^2} to get the final result.

Complete step by step answer:
We have to evaluate the integral xeax2dx\int {x{e^{ - a{x^2}}}} dx.
Let I=xeax2dx(1)I = \int {x{e^{ - a{x^2}}}} dx - - - (1).
Let u=ax2(2)u = - a{x^2} - - - (2)
Differentiating both the sides with respect to xx, we get
dudx=2ax\Rightarrow \dfrac{{du}}{{dx}} = - 2ax
On cross multiplication, we get
xdx=du2a(3)\Rightarrow xdx = - \dfrac{{du}}{{2a}} - - - (3)
Substituting (2)(2) and (3)(3) in (1)(1), we get
I=eu×(du2a)\Rightarrow I = \int {{e^u}} \times \left( { - \dfrac{{du}}{{2a}}} \right)
As we can take constant out of integration. On rewriting, we get
I=12aeudu\Rightarrow I = - \dfrac{1}{{2a}}\int {{e^u}} du
As we know that exdx=ex+C\int {{e^x}} dx = {e^x} + C. Using this, we get
I=12aeu+C\Rightarrow I = - \dfrac{1}{{2a}}{e^u} + C
Substituting back u=ax2u = - a{x^2}, we get
I=12aeax2+C\therefore I = - \dfrac{1}{{2a}}{e^{ - a{x^2}}} + C

Therefore, integral xeax2dx\int {x{e^{ - a{x^2}}}} dx is (12aeax2+C)\left( { - \dfrac{1}{{2a}}{e^{ - a{x^2}}} + C} \right).

Additional information: Antidifferentiation or indefinite integration is the process of solving for an antiderivative and its opposite operation i.e., the process of finding a derivative is called differentiation. Antiderivative is generally used to evaluate the indefinite integral of any function but it can also be used to compute definite integrals by using the fundamental theorem of calculus. The linearity of integration basically removes the complex integrand into simpler ones and hence becomes easier in case of definite integrals where limits are provided by just substituting at the place of dependent variable to get the definite answer.

Note: Here, CC is the constant of integration. In a definite integral we get the value of integration constant by using the upper and lower limit of integration. Note, we can also use inverse trigonometric properties to find the antiderivative. Using the direct integration formula that is mentioned as identity can make the problem simple by just substituting those values.