If $f$ is a continuous function on $\Bbb{R}$, then prove that $\int_0^1 f(x)x^2 \, \mathrm{d}x = f(c)/3$ for some $c \in[0,1]$.

Question

I tried to solve this using Riemann sum. I get integral is $f(c)c^2$ for some $c$ in $[0,1]$. But I couldn't show that it is $f(k)/3$ for some $k$ in $[0,1]$. Could you help me? Thanks.

Apply the mean value theorem for integrals https://en.wikipedia.org/wiki/Mean_value_theorem to the case $g(x) = x^2$. — trancelocation, Apr 30 '19 at 11:56

score 3 · Accepted Answer · answered Apr 30 '19 at 11:28

3

Hint. Note that $$\min_{t\in [0,1]} f(t) \cdot \int_0^1 x^2\,dx\leq \int_0^1 f(x)x^2\,dx \leq \max_{t\in [0,1]} f(t) \cdot \int_0^1 x^2\,dx.$$

answered Apr 30 '19 at 11:28

Robert Z

145,942

Thanks,I got it. – Balasubramannyan S Apr 30 '19 at 12:44

If $f$ is a continuous function on $\Bbb{R}$, then prove that $\int_0^1 f(x)x^2 \, \mathrm{d}x = f(c)/3$ for some $c \in[0,1]$.

1 Answers1