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Integration Question: Differentiating a definite integral

ISITIEIW

New member
Nov 4, 2013
17
So the question is…Evaluate the following…
\(\displaystyle \frac{d}{dx} \left(\int _1^{x^2} \cos(t^2) \, dt \right)\)

I thought i could use the FTC on this because it states…
\(\displaystyle \frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)\)

but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem

Thanks :)
 
Last edited by a moderator:

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Re: Integration Question

Since the lower bound is still a constant, it doesn't change the method you use to evaluate this. It could be at 0,1,2 or any constant and still get the same result. You're going to integrate and get a constant and then take the derivative of that constant, so you end up with 0 regardless.

Here you'll need to be careful to use the chain rule when applying the FTOC. What progress have you made so far? :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Integration Question

So the question is…Evaluate the following…
d/dx(integration of 1 to x^2 of cos(t^2)dt)

I thought i could use the FTC on this because it states…
d/dx(integration of 0 to x of f(t)dt)

but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem

Thanks :)
We are given to evaluate:

\(\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt\)

Now, the anti-derivative form of the FTOC tells us:

\(\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=\frac{d}{dx}\left(F(h(x))-F(g(x)) \right)\)

On the right, applying the chain rule, we obtain:

\(\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=F'(h(x))h'(x)-F'(g(x))g'(x)\)

Since \(\displaystyle F'(x)=f(x)\), we may write:

\(\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f(h(x))h'(x)-f(g(x))g'(x)\)

Applying this to the given problem, there results:

\(\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=\cos\left(\left(x^2 \right)^2 \right)(2x)-\cos\left(\left(1 \right)^2 \right)(0)\)

Simplifying, we find:

\(\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=2x\cos\left(x^4\right)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Let us evaluate \(\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)\) where \(\displaystyle 0<a<x\)

Then since \(\displaystyle \int_0^x f(t)\, dt =\int_0^a f(t)\, dt+ \int_a^x f(t)\, dt \)

\(\displaystyle \int_a^x f(t)\, dt=-\int_0^a f(t)\, dt+\int_0^x f(t)\, dt \)

If we differentiate then since the first integral is independent of $x$ we have

\(\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)=\frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)\)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The "Leibniz formula", a generalization of the fundamental theorem of calculus, says
[tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x, \beta(x))- \frac{d\alpha}{dx}f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt[/tex]