# Differentiate the given integral

• chwala
In summary, the integral $$\int_{x^2}^{2x} \sin t \, dt$$ is equal to $$-2x \sin x^2 +2 \sin 2x$$ using the fundamental theorem of calculus. This can be confirmed by differentiating the integral with respect to x and using the chain rule, resulting in the same expression.
chwala
Gold Member
Homework Statement
This is my own question (refreshing on the fundamental laws of calculus).

Differentiate the following integral:

$$\int_{x^2}^{2x} \sin t \, dt$$
Relevant Equations
calculus
My take:

$$\int_{x^2}^{2x} \sin t \, dt$$

using the fundamental theorem of calculus we shall have,

$$\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

I also wanted to check my answer, i did this by,

$$\int [-2x \sin x^2 +2 \sin 2x] dx$$

for the integration of the first part i.e

$$\int -2x \sin x^2 dx$$

i let ##u=x^2## giving me,

$$\int -\sin u du=\cos u+k = cos (x^2)+k$$and for the integration of the second part i.e

$$\int 2 \sin 2x dx$$

i let ##u=2x##, giving me,

$$\int 2 \sin 2x dx=\int -\sin u du=-\cos u +k=-\cos 2x+k$$

thus,

$$\int -2x \sin x^2 +2 \sin 2x dx=\cos x^2-\cos 2x+k$$

i just checked with wolframalpha and confirmed that:

$$\int_{x^2}^{2x} \sin t \, dt=\cos x^2-\cos 2x$$

I may get more insight or different ways of proof from you guys.

Last edited:
chwala said:
Homework Statement: This is my own question (refreshing on the fundamental laws of calculus).

Differentiate the following integral:

##\int_{x^2}^{2x} sin t \, dt##

You have
Code:
\int_x^2^{2x} sin t \, dt
which is inherently ambiguous. LaTeX cannot tell if \int_a^b^c means \int_a^{b^c} producing $\int_a^{b^c}$ or \int_{a^b}^c producing $\int_{a^b}^c$ or any other possible nesting of sub/superscripts. I assume, from the absence of logarithms elsewhere in your post, that you mean $$\int_{x^2}^{2x} \sin t\,dt = \cos (x^2) - \cos 2x.$$ The answer to your question is that $$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)\,dt = b'(x)f(b(x)) - a'(x)f(a(x)).$$

Last edited by a moderator:
topsquark and chwala
pasmith said:
You have
Code:
\int_x^2^{2x} sin t \, dt
which is inherently ambiguous. LaTeX cannot tell if \int_a^b^c means \int_a^{b^c} producing $\int_a^{b^c}$ or \int_{a^b}^c producing $\int_{a^b}^c$ or any other possible nesting of sub/superscripts. I assume, from the absence of logarithms elsewhere in your post, that you mean $$\int_{x^2}^{2x} \sin t\,dt = \cos (x^2) - \cos 2x.$$ The answer to your question is that $$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)\,dt = b'(x)f(b(x)) - a'(x)f(a(x)).$$
@pasmith i amended that...looks like you responded when i was still editing. I am conversant with the rule though. Cheers.

@chwala, the Fundamental Theorem of Calculus has two parts, one of which is to find an antiderivative of the integrand and then to evaluate the antiderivative at the two integration limits. The other part of the FTC shows how to determine the derivative of an integral, which is what @pasmith used in his explanation. I suspect that whoever wrote the problem intended for you to do it the way pasmith did.

topsquark and chwala
Mark44 said:
@chwala, the Fundamental Theorem of Calculus has two parts, one of which is to find an antiderivative of the integrand and then to evaluate the antiderivative at the two integration limits. The other part of the FTC shows how to determine the derivative of an integral, which is what @pasmith used in his explanation. I suspect that whoever wrote the problem intended for you to do it the way pasmith did.
@Mark44 this is my own problem ...i was actually going through:

https://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx

It is quite clear as i made use of the theorem directly...and just checked my own working...only needed insight which you have gave. Cheers mate.

Last edited:
chwala said:
Differentiate the following integral:

$$\int_{x^2}^{2x} \sin t \, dt$$
using the fundamental theorem of calculus we shall have $$\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$
Your work was very confusing to me because you omitted the fact that you were differentiating that integral above. IOW, your equation above is incorrect.
Edited: I omitted the minus sign in the earlier post. The corrected version is below:
$$\int_{x^2}^{2x} \sin t \, dt = \left. -\cos(t)\right |_{x^2}^{2x} = -\cos(2x) + \cos(x^2)$$

What you should have written is this: $$\frac d{dx} \int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

Also, when you used the FTC's first part, you really should have shown the work in going from ##\frac d{dx} \int_{x^2}^{2x} \sin t \, dt## to the expression you found, as you need to use the chain rule twice.

Last edited:
topsquark and erobz
chwala said:
Homework Statement: This is my own question (refreshing on the fundamental laws of calculus).

Differentiate the following integral:

$$\int_{x^2}^{2x} \sin t \, dt$$
Relevant Equations: calculus

My take:

$$\int_{x^2}^{2x} \sin t \, dt$$

using the fundamental theorem of calculus we shall have,

$$\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

I think you mean:

$$\frac{d}{dx}\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

Mark44 said:
Your work was very confusing to me because you omitted the fact that you were differentiating that integral above. IOW, your equation above is incorrect.
Edited to include the minus signs I omitted:
$$\int_{x^2}^{2x} \sin t \, dt = \left. -\cos(t)\right |_{x^2}^{2x} = -\cos(2x) + \cos(x^2)$$

What you should have written is this: $$\frac d{dx} \int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

Also, when you used the FTC's first part, you really should have shown the work in going from ##\frac d{dx} \int_{x^2}^{2x} \sin t \, dt## to the expression you found, as you need to use the chain rule twice.

Are you sure my equation is incorrect?

Last edited by a moderator:
erobz said:
I think you mean:
Absolutely; that is what i mean.

chwala said:
Are you sure my equation is incorrect?
As it is written. You have not shown the derivative out in front of the integral, that is what we are pointing out. You have written a statement which is false. Your workings for the problem are fine, your presentation of it is flawed.

Mark44, topsquark and chwala
My question is in reference to the integration by @Mark44 . Just check or read my post ##8## again.

...yes, my equation was incorrect as i missed out on the derivative... the other equation i am reffering to is the integration bit..i think a mistake there or typo. cheers guys.
erobz said:
As it is written. You have not shown the derivative out in front of the integral, that is what we are pointing out. You have written a statement which is false. Your workings for the problem are fine, your presentation of it is flawed.

Last edited:
chwala said:
My question is in reference to the integration by @Mark44 . Just check or read my post ##8## again.
Which integration?

SammyS said:
Which integration?
$$\int_{x^2}^{2x} \sin t \, dt = \left. \cos(t)\right |_{x^2}^{2x} = \cos(2x) - \cos(x^2)$$

chwala said:
Are you sure my equation is incorrect?
Yes, but mine was also incorrect, as I omitted the sign when I got the antiderivative of ##\cos(t)##. I fixed it in my earlier post and in the quoted posts of mine.

chwala said:
My question is in reference to the integration by @Mark44 .
chwala said:
$$\int_{x^2}^{2x} \sin t \, dt = \left. \cos(t)\right |_{x^2}^{2x} = \cos(2x) - \cos(x^2)$$
The above should be
$$\int_{x^2}^{2x} \sin t \, dt = \left. -\cos(t)\right |_{x^2}^{2x} = -\cos(2x) + \cos(x^2)$$
I have corrected this error of mine in the original post and in quoted posts.

As already mentioned, your error was in omitting the fact that you were differentiating the integral.

chwala
chwala said:
$$\int_{x^2}^{2x} \sin t \, dt = \left. \cos(t)\right |_{x^2}^{2x} = \cos(2x) - \cos(x^2)$$
That integration is nearly correct.

There is a sign error.

chwala
chwala said:
$$\int_{x^2}^{2x} \sin t \, dt = \left. \cos(t)\right |_{x^2}^{2x} = \cos(2x) - \cos(x^2)$$
What chwala quoted was my work, in which I omitted the sign on cos(t).
SammyS said:
That integration is nearly correct.

There is a sign error.
Right. I fixed it a little while ago.

chwala

• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
904
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
439
• Calculus and Beyond Homework Help
Replies
5
Views
802
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
566
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
617
• Calculus and Beyond Homework Help
Replies
11
Views
753