# an integration homework

Argh I had a picture for this but dont have enough reputation for anything on here. did it as I don't understand how to insert the maths notation in here.

If $f(x) = \int_0^{g(x)}(1+t^3)^{-1/2} dt$ where $g(x) = \int_0^{\cos x}(1+\sin (t^2)) dt$

I have to find $f'(\pi/2)$

I know it has something to do with substitution and I've tried integrating by parts and things like that too but its not working out. I think theres something about the way that the questions been asked thats not helping anyway thats the first question.

there'll be more in the future im sure.

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Put in the link to your picture. people here will take care. And, MathJaX is in place here. Enclose math between $...$ and get guided by your intuition for what a symbol could be! – user21436 Feb 24 '12 at 20:58
okay thanks for the advice. I'll play guesswork then. thanks henning for adding that in for me – Tessa Danger Bamkin Feb 24 '12 at 21:04

To make the work clear, put

$$h(x) =\int_0^{x}(1+t^3)^{-1/2} dt$$

$$m(x) = \int_0^{ x}(1+\sin (t^2)) dt$$

$$f(x) = \int_0^{g(x)}(1+t^3)^{-1/2} dt$$

$$g(x) = \int_0^{\cos x}(1+\sin (t^2)) dt$$

Then, we have that

$$F(x) = h \circ m \circ \cos(x)$$

Because of the chain rule we can solve this as follows:

$$F'(x) = h' \circ m \circ \cos(x) \cdot (m \circ \cos (x))'$$

$$F'(x) = h' \circ m \circ \cos(x) \cdot (m' \circ \cos (x)) \cdot (\cos (x))'$$

$$F'(x) = h' \circ m \circ \cos(x) \cdot (m' \circ \cos (x)) \cdot (-\sin(x))$$

So we now calculate our derivatives:

$$h'(x) = (1+x^3)^{-1/2}$$

$$m'(x) =1+\sin(x^2)$$

Now we're ready to plug in $\pi/2$. We have

$$-\sin (\pi/2)=-1$$

$$\cos (\pi/2) = 0$$

so that

$$m'(0) = 1+\sin (0) = 1$$

So for the last one we have

$$h' \circ m(0) = h'(0) = 1$$

Finally, we get

$$F'(\pi/2) = -1$$

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okay think i get it now thanks :). I just need to work on understanding chain rule as a composite functio I guess and not in the leibniz notation that I was first taught. – Tessa Danger Bamkin Feb 24 '12 at 21:56

Note that $g(x) = \tilde{g}(y) = \int_0^y(1+\sin (t^2)) dt$, where $y=\cos(x)$. Moreover $f(x)=\tilde{f}(z) =\int_0^z(1+t^3)^{-1/2} dt$, where $z=\tilde{g}(y)$. So $f(x) = \tilde{f}(\tilde{g}(\cos(x))) = (\mathrm{notation}) = \tilde{f}(z(y(x)))$.

$$f' = \frac{df}{dx} = \frac{d\tilde{f}(z(y(x)))}{dx} = \frac{d\tilde{f}}{dz} \cdot \frac{dz}{dy} \cdot \frac{dy}{dx} = (1 + z^3)^{-1/2} \cdot (1 + \sin(y^2)) \cdot (-\sin(x)) = (1 + 0^3)^{-1/2} \cdot (1+0) \cdot (-1) = -1$$ (note, that $z=g(x)=\tilde{g}(\cos(\pi/2))$ is an integral from $0$ to $0$, so $z=g(x)=0$).

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I dont really understand where the method comes from. it seems to me that you have multiplied both the main equations and differentiated one of the limits but im not sure why or what rule you're using. – Tessa Danger Bamkin Feb 24 '12 at 21:35
@TessaDangerBamkin: I rewrited that, is it clearer? I only use the chain rule and the fact that $d(\int_a^x f(t)dt)/dx = f(x)$. – savick01 Feb 24 '12 at 21:45

Write $H(t)$ for the first integrand.

If $f(x) = \int_0^{g(x)} H(t) \,dt$, then $f'(x)$ is defined as the limit of $(f(x+h)-f(x))/h$, for $h\to 0$, so you have to compute $$\lim_{h\to 0} \frac1h \int_{g(x)}^{g(x+h)} H(t)\,dt$$ The integral is "difference between bounds" times "average value"; for continuous $H$ and small $h$, the average value is close to $H(g(x))$. So the integral is close to $(g(x+h)-g(x)) \cdot H(g(x))$. The whole limit then comes out to $$\lim_{h\to 0} \frac1h (g(x+h)-g(x)) \cdot H(g(x))= g'(x)\cdot H(g(x)).$$ By the same reasoning, $g'(x) = (\cos x)'\cdot (1+\sin(\cos(x)^2))$.

Now plug in $x=\pi/2$, to get $g'(\pi/2) = (-1) \cdot (1+\sin(0)) = -1$.

Multiply this by $H(g(\pi/2))$. $\cos(\pi/2)=0$, so $g(\pi/2) = 0$, and $H(g(\pi/2)) = H(0) = 1$.

So the result, if I have not miscalculated, is $-1$.

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