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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 35 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 35 textbook solution.

Answers (1)

Answer:-  \frac{2}{\pi}

Hints:-  You must know the rules of integration.

Given:-  \int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x

Solution : f(x)=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x

|f(x)|=f(x) \text { when } x \geq 0 \text { and }-f(x) \text { when } x<0

\text { When } f(x)=x \cos (\pi x) \text {, we have }

\begin{aligned} &|x \cos (\pi x)|=\left\{\begin{array}{l} x \cos (\pi x) ;-1 \leq x \leq 1 / 2 \\ -x \cos (\pi x) ;-1 / 2 \leq x<0 \\ x \cos \pi x ; 0 \leq x<1 / 2 \\ -x \cos (\pi x) ; 1 / 2 \leq x<1 \end{array}\right. \\ &I=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x \end{aligned}

\\I=\int_{-1}^{-1 / 2} x \cdot \cos (\pi x)-\int_{-1 / 2}^{0} x \cdot \cos (\pi x) \cdot d x+\int_{0}^{1 / 2} x \cdot \cos (\pi x) \cdot d x-\int_{-1 / 2}^{1} x \cdot \cos (\pi x) \cdot d x

Now we can easily compile indefinite integral

\begin{aligned} I &=\int x \cdot \cos (\pi x) \cdot d x \\ &=\frac{x \cdot \cos (\pi x)}{\pi}+\frac{\cos (\pi x)}{\pi} \\ \end{aligned}

\therefore I^{\prime}=\left(I_{1 / 2}-I_{-1}\right)-\left(I_{0}-I_{-1 / 2}\right)+\left(I_{1 / 2}-I_{0}\right)-\left(I_{1}-I_{1 / 2}\right)

Where

\begin{gathered} I_{-1}=\frac{-1}{\pi^{2}} \\ I_{-1 / 2}=\frac{1}{2 \pi} \\ I_{0}=\frac{1}{\pi^{2}} \end{gathered}

\begin{aligned} &I_{1 / 2}=\frac{1}{2 \pi} \\ &I_{1}=\frac{-1}{\pi^{2}} \end{aligned}

Putting values

We get I=\frac{2}{\pi}

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