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\frac{d}{dt}\int_0^{\pi t^3}\frac{sin(x)}{x}dx

I got the answer to be \frac{3sint^3}{t} but I'm not sure that's right, I think I've missed the pi out somewhere.

I found \frac{dF}{dt}=3t^2F'(t^3) and F'(t^3)=\frac{sint}{t}
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Zhen Lin
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The fundamental theorem of calculus states that \displaystyle \frac{d}{dt}\int_{a}^{t} {f(x) \, dx} = f(t). Applying the chain rule: \displaystyle \frac{d}{dt}g(t) \cdot \frac{d}{d(g(t))}\int_{a}^{g(t)} {f(x) \, dx} = \cdots?
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student271828183
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(Original post by ......?)
\frac{d}{dt}\int_0^{\pi t^3}\frac{sin(x)}{x}dx

I got the answer to be \frac{3sint^3}{t} but I'm not sure that's right, I think I've missed the pi out somewhere.

I found \frac{dF}{dt}=3t^2F'(t^3) and F'(t^3)=\frac{sint}{t}
Let F(x)=\int\frac{sin(x)}{x}dx, we have
\frac{d}{dt}\int_0^{\pi t^3}\frac{sin(x)}{x}dx
=\frac{d}{dt}F(x)\mid_{o}^{\pi\t  ^3}
=\frac{d}{dt}F(\pi\ t^3)
=3\pi\ t^2f(\pi\ t^3)
=3\pi\ t^2\frac{sin(\pi\ t^3)}{\pi\ t^3}
=\frac{3sin(\pi\ t^3)}{t}
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