# I don't understand how to differentiate integrals

• sarahaha44
In summary, the OP is struggling with differentiation of definite integrals. Differentiation via the chain rule gives: \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x)-f(g(x))g'(x). For definite integrals, the next step is to apply the limits.
sarahaha44
I am in Calculus 2 and we're just reviewing calc 1. Can someone break down the concept of differentiating definite integrals for me? I am mostly struggling on the trig functions. The problem I am stuck on is
∫$^{}x$$_{}0$ cos(t$^{}2$) dt

You exploit the fact that the integration is the inverse of differentiation, so:

$$\int \frac{d}{dt}y(t)dt = y(t) + c$$ and $$\frac{d}{dt}\int y(t)dt = y(t) + c$$... for indefinite integrals. For definite integrals, the next step is to apply the limits.

But it looks like you want something a but different ... the chain rule for integrating $$\int x_0\cos(t^2)dt$$ ... is that correct?

If you want to evaluate: $$\int f(g(x))g^\prime(x).dx$$ then you can put ##u=g(x)## then solve to find out what dx is. When you make the substitution, you end up evaluating the integral of f(u)du which may be easier to do. That what you are after?

sarahaha44 said:
I am in Calculus 2 and we're just reviewing calc 1. Can someone break down the concept of differentiating definite integrals for me? I am mostly struggling on the trig functions. The problem I am stuck on is
∫$^{}x$$_{}0$ cos(t$^{}2$) dt
I think this is what you were trying to write:
$$\int_0^x cos(t^2)dt$$

Click what I wrote to see my LaTeX script, or click Quote on my post.

Simon Bridge said:
You exploit the fact that the integration is the inverse of differentiation, so:

But it looks like you want something a but different ... the chain rule for integrating $$\int x_0\cos(t^2)dt$$ ... is that correct?
I don't think so. I believe that 0 and x were the lower and upper limits of integration.

I don't think so. I believe that 0 and x were the lower and upper limits of integration.
Yeah that makes sense... we'll soon see :)

That particular example looks a tad rough for level 2 - have they just learned Fresnel integrals or something?

All you need to do is exploit the fundamental theorem of calculus:

$$\int_{g(x)}^{h(x)} f(t) dt = F(h(x))-F(g(x))$$
So differentiating via the chain rule gives:

$$\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x)-f(g(x))g'(x)$$
In the OP's example g(x)=0 and h(x)=x, so:

$$\frac{d}{dx} \int_{0}^{x} cos(t^2) dt = cos(x^2)-0$$

Mark44 said:
I think this is what you were trying to write:
$$\int_0^x cos(t^2)dt$$

Click what I wrote to see my LaTeX script, or click Quote on my post.

I don't think so. I believe that 0 and x were the lower and upper limits of integration.
Yes! This is what I was trying to write - sorry, new to the site :)

elfmotat said:
All you need to do is exploit the fundamental theorem of calculus:

$$\int_{g(x)}^{h(x)} f(t) dt = F(h(x))-F(g(x))$$
So differentiating via the chain rule gives:

$$\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x)-f(g(x))g'(x)$$
In the OP's example g(x)=0 and h(x)=x, so:

$$\frac{d}{dx} \int_{0}^{x} cos(t^2) dt = cos(x^2)-0$$
Ah thank you! Life makes sense again haha

elfmotat said:
All you need to do is exploit the fundamental theorem of calculus:

$$\int_{g(x)}^{h(x)} f(t) dt = F(h(x))-F(g(x))$$
So differentiating via the chain rule gives:

$$\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x)-f(g(x))g'(x)$$
In the OP's example g(x)=0 and h(x)=x, so:

$$\frac{d}{dx} \int_{0}^{x} cos(t^2) dt = cos(x^2)-0$$
Actually I am unclear with your last step there. So f(h(x))h'(x) equals x, do you just plug x in for t? What if it equaled x2? Would it end up being cos x4? Hypothetically speaking.

sarahaha44 said:
Actually I am unclear with your last step there. So f(h(x))h'(x) equals x, do you just plug x in for t? What if it equaled x2? Would it end up being cos x4? Hypothetically speaking.

If you were integrating from 0 to x2, you would get:

$$2xcos(x^4)$$

The "fundamental theorem of calculus" (part of it) says the if $F(x)= \int f(x) dx$, then dF/dx= f(x). That says that $d/dx \int_a^x f(t)dt= f(x)$. (The other part says that if f(x)= dF/dx, then $F(x)= \int f(x)dx+ C$ for some constant C.)

"Leibniz's formula" is more general:
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt=f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x, \alpha(x))\frac{d\alpha(x)}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, t)}{\partial x}dt$$.

The "fundamental theorem" is clearly a special case of "Leibniz's formula" with $\alpha(x)= 0$, so that $d\alpha/dx= 0$, $\beta(x)= x$ so that $d\beta/dx= 1$ and f(t) not a function of x so that $\partial f/\partial x= 0$:
$$\frac{d}{dx}\int_0^x f(t)dt= (1)f(x)- (0)f(0)+ \int_0^x 0 dt= f(x)$$.

To do the initial example, $d/dx\int_0^x cos(t^2)dt$, you just need the "fundamental theorem", $d/dx\int_0^x cos(t^2)dt= cos(x^2)$.

To do the example elfmotat gives, $d/dx \int_0^{x^2} cos(t^2)dt$ take $\beta(x)= x^2$, so that $d\beta/dx= 2x$, and $\alpha(x)= 0$. $f(x, t)= cos(t^2)$ which does NOT depend on x so that $\partial f/\partial x= 0$. Leibniz' forumula becomes
$$2x cos((x^2)^2)- 0 cos(0^2)+ \int_0^{x^2} 0 dt= 2x cos(x^4)$$.

## 1. What is the purpose of differentiating integrals?

Differentiating integrals is a common technique used in calculus to find the rate of change of a function. It allows you to determine how quickly a function is changing at a particular point, which can be useful in a variety of real-world applications.

## 2. How is differentiating integrals different from finding derivatives?

Differentiating integrals is essentially the same as finding derivatives, but with the added step of first integrating the function. This means that you must use the rules of integration to find the antiderivative of the function before applying the rules of differentiation.

## 3. Can you explain the process of differentiating integrals?

The process of differentiating integrals involves using the rules of integration to find the antiderivative of the function, and then applying the rules of differentiation to the resulting function. This will give you the derivative of the original function.

## 4. What are some common techniques used in differentiating integrals?

Some common techniques used in differentiating integrals include the power rule, product rule, quotient rule, and chain rule. These rules can be used to differentiate a wide variety of functions.

## 5. How can I improve my understanding of differentiating integrals?

To improve your understanding of differentiating integrals, it is important to practice and familiarize yourself with the various rules and techniques. You can also seek out additional resources, such as textbooks or online tutorials, to gain a deeper understanding of the concepts and applications of differentiating integrals.

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