Question

# Evaluate the definite integral $\int_0^1 {x{e^{{x^2}}}dx}$

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Hint: n mathematics, integration is the concept of calculus and it is the act of finding the integrals. Here we will find integration, by using the concept of equivalent value, where the same term will be multiplied and divided and will simplify and then will place the formula and simplify for the resultant answer. Here we will assume first for the part of integration, and then find its derivative and the function of the given integral with respect to “x” and then will place the limit in the function and then simplify it for the resultant value.

Let us assume that $F(x) = \int_0^1 {x{e^{{x^2}}}dx}$ .... (A)
Suppose ${x^2} = t$
Differentiate with respect to “x” on both the sides of the equation-
$\dfrac{{d({x^2})}}{{dx}} = \dfrac{{dt}}{{dx}}$
Apply identity $\dfrac{d}{{dx}}{x^n} = n.{x^{n - 1}}$ in the above equation –
$\Rightarrow 2x = \dfrac{{dt}}{{dx}}$
Do Cross-multiplication in the above equation, where numerator of one side is multiplied with the denominator of the opposite side.
$\Rightarrow dx = \dfrac{{dt}}{{2x}}$
Place above value in the equation (A)
$\int_{}^{} {x{e^{{x^2}}}dx = } \int_{}^{} x .{e^t}\dfrac{{dt}}{{2x}}$
Take constant common from the denominator and re-write the equation -
$\int_{}^{} {x{e^{{x^2}}}dx = } \dfrac{1}{2}\int_{}^{} x .{e^t}\dfrac{{dt}}{x}$
Common factors from the numerator and the denominator cancel each other. Therefore remove “x” from the numerator and the denominator.
$\int_{}^{} {x{e^{{x^2}}}dx = } \dfrac{1}{2}\int_{}^{} {{e^t}} dt$
Using identity –
$\int_{}^{} {x{e^{{x^2}}}dx = } \dfrac{1}{2}{e^t}$
Now, replace $t = {x^2}$ in the above equation –
$\int_{}^{} {x{e^{{x^2}}}dx = } \dfrac{1}{2}{e^{{x^2}}}$
Now, if $F(x) = \dfrac{1}{2}{e^{{x^2}}}$
Then – $\int_0^1 {x{e^{{x^2}}}dx}$ can be expressed as
$\int_0^1 {x{e^{{x^2}}}dx} = F(1) - F(0)$
Place the value for “x” in the above equation –
$\int_0^1 {x{e^{{x^2}}}dx} = \dfrac{1}{2}{e^{{1^2}}} - \dfrac{1}{2}{e^{{0^2}}}$
Exponential raise to zero is 1, ${e^0} = 1$ place in the above equation.
$\int_0^1 {x{e^{{x^2}}}dx} = \dfrac{1}{2}{e^1} - \dfrac{1}{2}$
Take a common multiple outside the bracket.
$\int_0^1 {x{e^{{x^2}}}dx} = \dfrac{1}{2}\left( {e - 1} \right)$
So, the correct answer is “ $\int_0^1 {x{e^{{x^2}}}dx} = \dfrac{1}{2}\left( {e - 1} \right)$ ”.

Note: The difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverse of each other.