How to solve a definite integral?
If $\int\limits_0^9f(x)\,dx=5$, find $\int\limits_0^3f(3x)\,dx$.
I know how to solve integrals and definite integrals, but I am totally confused on what to do in this problem. I have no idea where to start. Any suggestions would be helpful! Thank you!
$\endgroup$ 24 Answers
$\begingroup$Let $u=3x$, then $du=3dx$ and$$\int_0^3f(3x)dx=\int_0^9f(u)\frac{du}{3}$$Thus $$\int_0^3f(3x)dx=\frac{1}{3}\int_0^9f(x)dx=\frac{5}{3}$$
$\endgroup$ $\begingroup$Let $t=3x$, then:
$$\int\limits_0^3f(3x)\,dx={1\over 3}\int\limits_0^9f(t)\,dt=...$$
$\endgroup$ $\begingroup$Substitute $t=3x$. Then $\int_{0}^{3}f(3x)dx=\frac{1}{3}\int_0^9f(t)dt=\frac{5}{3}$.
$\endgroup$ $\begingroup$$f(3x)$ is a stretch of $f(x)$ by scale factor $1/3$ in the x-direction. So if the integral of $f(x)$ between 0 and k is the sum if the areas of rectangles, then the integral of $f(3x)$ between 0 and $k/3$ is also the sum of some rectangles - the same height as the ones for $f(x)$, but with each rectangle having $1/3$ the width
$\endgroup$
