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If $f$ is a differentiable function on the interval $[0,4]$, find $$ \int_{0}^{2} tf'(t^2)\,\mathrm{d}t.$$ For which $f$ is the integral positive?

For the above question, I integrated using integration by parts. I let $u=t$ and $\frac{\mathrm{d}v}{\mathrm{d}t}=f'(t^2)$ so that $\frac{\mathrm{d}u}{\mathrm{d}t}=1$ and $v=f(t^2)$. Then using integration by parts formula I get $$ I=uv-\int\left(v\frac{\mathrm{d}v}{\mathrm{d}t}\right)\mathrm{d}t$$ where $uv=tf(t^2)$ and I don't know how to simply the integral part? Is integration by parts even required for this question or do I have to adopt a completely different approach?

Any help would be much appreciated.

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First off, if you take $$ \frac{\mathrm{d}v}{\mathrm{d}t} = f'(t^2), $$ You will not get $v = f(t^2)$. You can check this by differentiating: $$ \frac{\mathrm{d}}{\mathrm{d}t} f(t^2) = f'(t^2) \frac{\mathrm{d}}{\mathrm{d}t} t^2 = f'(t^2) 2t.$$ Remember the chain rule!

However, if instead of attempting to integrate by parts, you hit the integral with a change of variables, you will obtain \begin{align} \int_{0}^{2} t f'(t^2)\mathrm{d}t &= \int_{0}^{4} f'(u)\,\frac{\mathrm{d}u}{2} &&\text{(let $u=t^2$, so that $\mathrm{d}u = 2\mathrm{d}t$)} \\ &= \frac{1}{2} \int_{0}^{4} f'(u)\,\mathrm{d}u \\ &= \frac{1}{2} \big( f(4) - f(0) \big). &&\text{(by the FTC)} \end{align} Thus $$ \int_{0}^{2} t f'(t^2)\mathrm{d}t > 0 \iff \frac{1}{2} \big( f(4) - f(0) \big) > 0 \iff f(4) > f(0),$$ which gives the desired result.

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Let $u = t^2$, then $du = 2t dt$, hence this integral is $\frac{1}{2}\int_0^4 f'(u) du = (f(4) - f(0))/2$. Therefore the integral is positive iff $f(4) > f(0)$.

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