Question

Evaluate the definite integral $\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)}dx$

Answer 1

Break the terms of the given integral into two parts. Calculate the integral of $\sin \dfrac{\pi x}{4}$ by using the formula $\int{\sin \left( ax+b \right)dx=\dfrac{-\cos \left( ax+b \right)}{a}}$ , here a and b are the constants. Now to calculate the integral of $x{{e}^{x}}$ , assume x as function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and ${{e}^{x}}$ as function 2 $\left( {{f}_{2}}\left( x \right) \right)$ and apply the rule of integration by parts given as$\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ to get the answer. Here, ${{f}_{1}}'\left( x \right)=\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right) \right)$.

Complete step-by-step solution:
Here, we are required to evaluate the integral $\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)}dx$. Let us assume its value as I. so we have,
$\Rightarrow I=\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)}dx$
Breaking the terms, we get
$\Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\int\limits_{0}^{1}{\sin \dfrac{\pi x}{4}dx}$
Now we know that $\int{\sin \left( ax+b \right)dx=\dfrac{-\cos \left( ax+b \right)}{a}}$, where a and b are constants. So, we get

& \Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\left[ \dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi }{4}} \right]_{0}^{1} \\\ & \Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\dfrac{4}{\pi }\left[ -\cos \dfrac{\pi x}{4} \right]_{0}^{1} \\\ \end{aligned}$$ Substituting the limits, we get $$\begin{aligned} & \Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\dfrac{4}{\pi }\left[ -\cos \dfrac{\pi }{4}-(-\cos 0) \right] \\\ & \Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\dfrac{4}{\pi }\left[ -\dfrac{1}{\sqrt{2}}+1 \right] \\\ & \Rightarrow I=\int\limits_{0}^{1}{x{{e}^{x}}dx}+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right]\ldots \ldots \ldots \left( i \right) \\\ \end{aligned}$$ Now, we have to calculate the integral of $x{{e}^{x}}$ which is a product of two functions. That means we have to apply the integration by parts. Clearly, we can see that in the product $x{{e}^{x}}$, x is algebraic and ${{e}^{x}}$ is an exponential function. So, according to the ILATE rule we have to assume x as the function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and ${{e}^{x}}$ as function 2 $\left( {{f}_{2}}\left( x \right) \right)$. Here, ILATE stands for I – Inverse trigonometric function L – Logarithmic function A – Algebraic function T – Trigonometric function E – Exponential function So, the numbering of the functions is done according to the order of appearance in the above list. Now to calculate the integral apply the formula $\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ and substitute the limits. Here, $${{f}_{1}}'\left( x \right)=\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right) \right)$$. So applying the formula for $x{{e}^{x}}$ in equation (i), we get $$\begin{aligned} & \Rightarrow I=\left[ x\int{{{e}^{x}}dx} \right]_{0}^{1}-\int\limits_{0}^{1}{\dfrac{dx}{dx}\left( \int{{{e}^{x}}dx} \right)dx}+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ & \Rightarrow I=\left[ x{{e}^{x}} \right]_{0}^{1}-\int\limits_{0}^{1}{{{e}^{x}}dx}+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ & \Rightarrow I=\left[ x{{e}^{x}} \right]_{0}^{1}-\left[ {{e}^{x}} \right]_{0}^{1}+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ \end{aligned}$$ Substituting the limits, we get $$\begin{aligned} & \Rightarrow I=\left[ 1{{e}^{1}}-0{{e}^{0}} \right]-\left[ {{e}^{1}}-{{e}^{0}} \right]+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ & \Rightarrow I={{e}^{1}}-\left[ {{e}^{1}}-1 \right]+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ & \Rightarrow I=e-e+1+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ & \Rightarrow I=1+\dfrac{4}{\pi }\left[ 1-\dfrac{1}{\sqrt{2}} \right] \\\ \end{aligned}$$ **Note:** One may note that whenever we have to calculate the integral of products of two different functions we have to apply integration by parts method and numbering should be done according to the ILATE rule. So, you must remember the ILATE rule and the formula used for the integration by parts method to solve the above questions. Remember the basic formulas of integration of functions like ${{e}^{x}},\sin x,\cos x$ etc…