Question

Integrals question(poste on comment section)

Answer 1

@ganeshie8

Answer 2

i aasume you take the integral and then solve for the derivative?

Answer 3

There's a much simpler way. Recall part II of the FTC: \(\displaystyle\frac{d}{dx}\int_a^xf(t)\,dt = f(x)\). In this case, \(f(t) = (t+1)\sqrt{t}\).

So what do you get when you apply this to your problem? :-)

Answer 4

give me a sec

Answer 5

woops, \(\large f(t) = (t+1)\sqrt[3]{t}\); sorry about that!

Answer 6

I get

Answer 7

thats the derivative of that

Answer 8

is that the answer? @ChristopherToni

Answer 9

Unfortunately it isn't. When applying the formula I provided above, you don't need to differentiate the integrand. And since the upper limit of the integral is x and the lower limit is a constant (-2 in this case), you just substitute x into your f(t). Applying the formula properly should leave you with \(\large \displaystyle\frac{d}{dx}\int_{-2}^x (t+1)\sqrt[3]{t}\,dt = (x+1)\sqrt[3]{x}\).

Does this make sense?

Answer 10

@ChristopherToni i get it so then i would solver for the derivative?

Answer 11

But with the FTC formula, \(\large (x+1)\sqrt[3]{x}\) is the derivative. You're done after applying that, so you don't need to do anything else; that's why it's so nice to use! lol

Answer 12

ohh lol

Answer 13

The reason why this may look confusing is because you defined F(x) as an integral. xD

Answer 14

@ChristopherToni and for this one would be the same or different?

Answer 15

Very similar, but this tine note where the variable in the limit is: it's at the bottom. In the formula I provided you, you need to have the variable limit at the top. So you first want to rewrite \(\large \displaystyle\int_t^2 \frac{e^x}{x}\,dx\) as \(\large \displaystyle -\int_2^t\frac{e^x}{x}\,dx \) and then when you differentiate the integral, just substitute t into the integrand. What do you get when you do it this time?

Answer 16

time*

Answer 17

i know how to differentiate but how do i do that one?

Answer 18

You just substitute t into the integrand. So if \(\large \displaystyle F(t) = -\int_2^t\frac{e^x}{x}\,dx\), then \(\large \displaystyle F^{\prime}(t) = \frac{d}{dt}\int_2^t \frac{e^x}{x}\,dx = \dfrac{e^t}{t}\) and you're done!

Does this make sense?

Answer 19

(forgot the negative; it should be \(F^{\prime}(t) = -\dfrac{e^t}{t}\))

Answer 20

More or less, the reason why we use this formula from the FTC is due to the fact that you can't always find the antiderivative of the integrand provided, like seen in this case.

Answer 21

thanks i got it @ChristopherToni