ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 7.2 definite integral

For any real number \(x\), let \([ x ]\) denote the largest integer less than equal to \(x\). Let \(f\) be a real valued function defined on the interval \([-10,10]\) by \(f(x)=\left\{\begin{array}{cl}x-[x], & \text { if }(x) \text { is odd } \\ 1+[x]-x & \text { if }(x) \text { is even }\end{array}\right.\) Then the value of \(\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x\) is.

  1. A \(4\)
  2. B \(2\)
  3. C \(1\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(4\)

Step-by-step Solution

Detailed explanation

\(f ( x )\) is periodic function whose period is 2 \(\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x=\frac{\pi^{2}}{10} \times 10 \int_{0}^{2} f(x) \cos \pi x d x\) \(=\pi^{2}\left(\int_{0}^{1}(1-x) \cos \pi x d x+\int_{1}^{2}(x-1) \cos \pi x d x\right)\) Using by parts…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app