Simple Integration and Differentiation Problem

[Peter G.]

The question is the following:

y = (lower limit 0, upper limit e^2x)∫ (1/T)

Find y'(2)

According to the markscheme, the answer is equal to 4e². I got 2e².

I did the following:

I integrated the expression, which yielded: 2√t. I then substituted the upper and lower limits in, which gave me:

2e^x

I then differentiated that, which yields:

2e^x

I proceeded to substitute 2 in, which gave me 2e²

Can anyone shed some light onto this for me please?

Thank you in advance!

 

[Bandarigoda]

Could u write it on a paper and post. I'm on phone, can't read the equation clearly. If I could I will try help

 

[STEMucator]

The question is the following:

y = (lower limit 0, upper limit e^2x)∫ (1/T)

Find y'(2)

According to the markscheme, the answer is equal to 4e². I got 2e².

I did the following:

I integrated the expression, which yielded: 2√t. I then substituted the upper and lower limits in, which gave me:

2e^x

I then differentiated that, which yields:

2e^x

I proceeded to substitute 2 in, which gave me 2e²

Can anyone shed some light onto this for me please?

Thank you in advance!

Are you familiar with the fundamental theorem of calculus? If so, what is the result of this :

$$\frac{d}{dx} y = \frac{d}{dx} \int_{0}^{e^{2x}} \frac{1}{t} dt$$

 

[Peter G.]

I forgot to add (and this might answer your question, Zondrina) that, my first instinct was the following: If y is equal to the integral of 1/t, then the derivative of y should be equal to 1/t. But, if I went on and substituted 1/2 into the equation, I would be left with 1/2, which is not equal to any of the options available.

Anyway, is the answer to what you asked me (1/T)?

 

[Mark44]

I forgot to add (and this might answer your question, Zondrina) that, my first instinct was the following: If y is equal to the integral of 1/t, then the derivative of y should be equal to 1/t.

You are really oversimplifying things. Here is what is true:
$$ \text{If } f(x) = \int_0^x \frac{dt}{t}$$
then f'(x) = 1/x

The fact that your integral has an upper limit of e^2x rather than x is very significant, and means that you will need to apply the chain rule when you differentiate.

But, if I went on and substituted 1/2 into the equation, I would be left with 1/2, which is not equal to any of the options available.

Anyway, is the answer to what you asked me (1/T)?

Also, do not delete the homework template when you post a problem. It's there for a reason.

 

[STEMucator]

An easy way to remember the theorem is this :

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$

It will give you the correct answer.

 

[Mark44]

An easy way to remember the theorem is this :

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$

It will give you the correct answer.

Speaking only for myself, I find it simpler to use the chain rule directly than to remember a canned formula (that could possibly be remembered incorrectly).

$$\frac{d}{dx} \int_a^{u(x)} f(t) dt =\frac{d}{du} \left( \int_a^{u(x)} f(t) dt \right) * \frac{du}{dx}$$
$$=f(u(x)) * \frac{du}{dx}$$

Applying this idea to the original integral entails splitting it into two integrals.

 

[Peter G.]

Thank you very much to both of you. As a matter of fact, I knew the "canned formula" but had never been introduced to the concept of chain rule through this way, hence, I had no idea I had to apply it here.

Thank you once again and sorry for deleting the template.