Question:
If \(f(x) + f(\pi-x) = \pi^{2}\) then \(_{0}\int^{\pi}f(x) sinx dx\) ?
If \(f(x) + f(\pi-x) = \pi^{2}\) then \(_{0}\int^{\pi}f(x) sinx dx\) ?
Updated On: Aug 21, 2024
- π2
- \(\frac{\pi_2}{4}\)
- 2π2
- \(\frac{\pi_2}{8}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): π2
\(I=\int_0^\pi f(x)sinxdx\)
\(I=\int_0^\pi f(\pi-x)sinxdx\)
\(2I=\int_0^\pi sin(x)(f(x)+f(\pi-x))dx\)
\(2I=\pi^2\int_0^\pi sinxdx\)
\(2I=2\pi^2\int_0^\pi sinxdx\)
\(I=\pi^2(-cos x)^\frac{\pi}{2}_0\)
\(I=\pi^2\)
\(I=\int_0^\pi f(x)sinxdx\)
\(I=\int_0^\pi f(\pi-x)sinxdx\)
\(2I=\int_0^\pi sin(x)(f(x)+f(\pi-x))dx\)
\(2I=\pi^2\int_0^\pi sinxdx\)
\(2I=2\pi^2\int_0^\pi sinxdx\)
\(I=\pi^2(-cos x)^\frac{\pi}{2}_0\)
\(I=\pi^2\)
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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:
