Question

An integral a day:
Day 4. Today, I've got something a little different, I came across it yesterday and thought it was really cool. \[\int_1^{x^2} f(t)dt = 2\sin(x)e^x\] find f(t).

Answer 1

using the fundamental theorem of calculus I get\[f(x^2)=2e^x(\cos x+\sin x)\frac{dx}{dt}\]I'm not sure that getting this differential equation helps us though...

Answer 2

you're on the right track, but you're missing something, and that dx by dt shouldn't be there.

Answer 3

what you wanna do is assume that there is some function that the indefinite integral of f(t) evaluates to, just call it F(t), it should be pretty straightforward from there i hope

Answer 4

Solution:
using the fundamental theorem of calculus we can let the indefinite integral of f(t) be F(t), in which case we can say that:\[\int_1^{x^2}f(t)dt = F(x^2)-F(1) \\ F(x^2)-F(1) = 2 \sin(x)e^x \] and now its just simple matter of reducing F(x^2) back to f(t), so for starters, we take the derivative of both sides with respect to x:\[\frac{d}{dx}(F(x^2)-F(1))=\frac{d}{dx}(2\sin(x)e^x) \\ \text{chain rule:}~~ 2xf(x^2) =2e^x(\cos(x)+\sin(x))\]solve for f(x^2) then substitute in t. \[f(x^2) = \frac{e^x(\cos(x)+\sin(x))}{x}\\ f(t) = \frac{e^\sqrt{t}(\cos(\sqrt{t}+\sin(\sqrt{t})}{\sqrt{t}}\]

Answer 5

more easy. Think about bouth sides like functions of x2 and diferentiate bouth sides. De diferential of the integral by it's upper limit is the f(t). The rest is the same like you