Question

Solve the following integral-
$\int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx$ (n is positive integer) is equal to
A) n!
B) (n-1)!
C) (n-2)!
D) (n+1)!

Answer 1

The integral of type $\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx$ is solved by using gamma function i.e.

n \,}} \right. $$ Here, are some standard result for gamma function $\left| \\!{\overline {\, n \,}} \right. $: $ \left| \\!{\overline {\, {n + 1} \,}} \right. = n! \\\ \left| \\!{\overline {\, {\dfrac{1}{2}} \,}} \right. = \sqrt \pi \\\ \left| \\!{\overline {\, 1 \,}} \right. = 1 \\\ $ **Complete step-by-step answer:** Step 1: Given integral $$I = \int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx$$. In order to use gamma function: $$\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx = \left| \\!{\overline {\, n \,}} \right. $$ Step 2: Reduce the given integral accordingly to use gamma function. Here, the power of the variable, x is in the form $(n - 1)$. Therefore, reduce the power of variable, x in given integral in the form $(n - 1)$. The power of variable, x in given integral = n Adding +1 and -1 to the power won’t change the function. $ \Rightarrow n - 1 + 1$ $ \Rightarrow \left[ {(n + 1) - 1} \right]$ Thus, the power of variable, x in given integral can be written as $\left[ {\left( {n + 1} \right) - 1} \right]$ Step 3: Solve integral using gamma function Hence, $$I = \int\limits_0^\infty {\mathop x\nolimits^{\left[ {\left( {n + 1} \right) - 1} \right]} } \mathop e\nolimits^{ - x} dx$$. By gamma function $$I = \int\limits_0^\infty {\mathop x\nolimits^{\left[ {\left( {n + 1} \right) - 1} \right]} } \mathop e\nolimits^{ - x} dx = \left| \\!{\overline {\, {n + 1} \,}} \right. $$ $ \Rightarrow \left| \\!{\overline {\, {n + 1} \,}} \right. = n!$ Final answer: The integral $$\int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx = n!$$. **Thus the correct option is (A).** **Note:** To use gamma function, the limits of the integral must be from $0 \to \infty $. The question of type $$\int\limits_0^\infty {\mathop e\nolimits^{ - \mathop x\nolimits^2 } } dx$$can also be solved by the help of gamma function. $$I = \int\limits_0^\infty {\mathop e\nolimits^{ - \mathop x\nolimits^2 } } dx$$ Take $$\mathop x\nolimits^2 = t$$ $\therefore x = \sqrt t $ On differentiating both sides $ 2xdx = dt \\\ \Rightarrow dx = \dfrac{{dt}}{{2x}} \\\ $ $ \Rightarrow dx = \dfrac{{dt}}{{2\sqrt t }} \\\ \Rightarrow dx = \dfrac{1}{2}\mathop t\nolimits^{ - \dfrac{1}{2}} {\text{ }}dt \\\ $ Change of limits: When, $x \to 0 \Rightarrow t \to 0$ $x \to \infty \Rightarrow t \to \infty $ On substitution integral becomes $I = \dfrac{1}{2}\int\limits_0^\infty {\mathop t\nolimits^{ - \dfrac{1}{2}} } \mathop e\nolimits^{ - t} {\text{ }}dt$ In order to use gamma function: $$\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx = \left| \\!{\overline {\, n \,}} \right. $$ Here, the power of the variable, x is in the form $(n - 1)$. Therefore, reduce the power of the variable, t in above integral in the form $(n - 1)$. The power of variable, t in above integral = $ - \dfrac{1}{2}$ Adding +1 and -1 to the power won’t change the function. $ \Rightarrow - \dfrac{1}{2} - 1 + 1$ $ \Rightarrow \left( {\dfrac{1}{2} - 1} \right)$ Thus, the power of variable, t in above integral can be written as $\left( {\dfrac{1}{2} - 1} \right)$ Hence, $I = \dfrac{1}{2}\int\limits_0^\infty {\mathop t\nolimits^{\left( {\dfrac{1}{2} - 1} \right)} } \mathop e\nolimits^{ - t} {\text{ }}dt$ By gamma function $ \Rightarrow \dfrac{1}{2}\left| \\!{\overline {\, {\dfrac{1}{2}} \,}} \right. $ We know, $\left| \\!{\overline {\, {\dfrac{1}{2}} \,}} \right. = \sqrt \pi $ $\because I = \dfrac{{\sqrt \pi }}{2}$