Question:

If \(f(x) + f(\pi-x) = \pi^{2}\) then \(_{0}\int^{\pi}f(x) sinx dx\) ?

Updated On: Aug 21, 2024
  • π2
  • \(\frac{\pi_2}{4}\)
  • 2
  • \(\frac{\pi_2}{8}\)
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The Correct Option is A

Solution and Explanation

The correct option is (A): π2
\(I=\int_0^\pi f(x)sinxdx\)
\(I=\int_0^\pi f(\pi-x)sinxdx\)
\(2I=\int_0^\pi sin(x)(f(x)+f(\pi-x))dx\)
\(2I=\pi^2\int_0^\pi sinxdx\)
\(2I=2\pi^2\int_0^\pi sinxdx\)
\(I=\pi^2(-cos x)^\frac{\pi}{2}_0\)
\(I=\pi^2\)
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Questions Asked in JEE Main exam

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Concepts Used:

Integration by Parts

Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

  • u is the first function u(x)
  • v is the second function v(x)
  • u' is the derivative of the function u(x)

The first function ‘u’ is used in the following order (ILATE):

  • 'I' : Inverse Trigonometric Functions
  • ‘L’ : Logarithmic Functions
  • ‘A’ : Algebraic Functions
  • ‘T’ : Trigonometric Functions
  • ‘E’ : Exponential Functions

The rule as a diagram: