How can I change the limit of $\displaystyle\int\limits_0^\pi f(x)\,dx$ to $\displaystyle\int\limits_0^{\pi/2} f(x)\,dx$ ?
Change the limit of an integral
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$\begingroup$
integration
definite-integrals
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0If you mean "how can I compute $\int^{\pi/2}_0f(x)dx$ from $\int^\pi_0f(x)dx$?", then you can't without some conditions on $f$. The behaviour of $f$ in the two halves of the interval could be completely unrelated. – 2012-10-16
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0@Matt Pressland ...no...i just want the way to change the limit( 0 to $\pi$ )to( 0 to $\pi/2$) – 2012-10-16
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2We have that $\int_0^\pi f(x)dx = 2\int_0^{\pi/2} f(2x)dx$, but without any other information on $f$ that's the best you can do. – 2012-10-16
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0@Arthur....My problem is to solve $\int_o^\pi Sin^4{x}\,dx $ – 2012-10-16
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3Since $\sin^4(x)$ is symmetric about $x=\pi/2$ we have $\int_0^\pi \sin^4(x)dx = 2\int_0^{\pi/2} \sin^4(x)$. – 2012-10-16
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0@Arthur...thanks... – 2012-10-16
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0@Arthur...If the function is $Sin^4({\pi}x/a)$ where "a" is constant then what will happen? – 2012-10-16
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0No. Because swapping $x$ with $\pi x/a$ will change the axis of symmetry. – 2012-10-16
1 Answers
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$\int_o^\pi f(x)dx = \int_0^{\pi/2} 2 f(2y)dy$
by the change of variables $y := x/2$. Is this what you are looking for?