How Do You Calculate f(8) Using the Fundamental Theorem of Calculus?

[B18]

Homework Statement

If f is continuous, f(3)=10, and ∫ from (3,8)f'(x)dx=16, find f(8)

The Attempt at a Solution

I attempted to do this by evaluating 16x] from 3 to 8 using FTC. I got 80. And in no way shape or form did i get any closer to the answer from there.

 

[SammyS]

Homework Statement

If F is continuous, f(3)=10, and ∫ from (3,8)f'(x)dx=16, find f(8)

The Attempt at a Solution

I attempted to do this by evaluating 16x] from 3 to 8 using FTC. I got 80. And in no way shape or form did i get any closer to the answer from there.

I suspect that you mean that f is continuous. There is no F mentioned anywhere else.

Isn't $\displaystyle \int f'(x)\,dx=f(x)+C\ ?$

 

[B18]

Yes i apologize, the question has "f is continuous" not F.

 

[SammyS]

Yes i apologize, the question has "f is continuous" not F.

In addition to that ...

My rhetorical question was meant to be a hint.

Try it.

 

[B18]

Would this be a correct way of finding the answer?
f(3)=10 so f(8)=10+C
then ∫(3,8) f'(x)dx=16
f(x)+c=16
c=16 so..
f(8)=10+16=26.

 

[SammyS]

Would this be a correct way of finding the answer?
f(3)=10 so f(8)=10+C
then ∫(3,8) f'(x)dx=16
f(x)+c=16
c=16 so..
f(8)=10+16=26.

Well, f(8) is 26, but I don't believe you can determine C from the information given.

According to the FTC: $\displaystyle \ \int_{3}^{8}f'(x)\,dx=f(8)-f(3)\ .$

That's about all you need.

 

[B18]

i did ∫(3,8)=f(8)-f(3)=16
f(8)=16+f(3)
f(8)=16+10=26
correct way of completing this problem?

 

[Mark44]

i did ∫(3,8)=f(8)-f(3)=16
f(8)=16+f(3)
f(8)=16+10=26
correct way of completing this problem?

The answer looks good.

Your notation could use some work though.

$$ \int_3^8 f'(x) dx = 16$$
Replacing the integral above, for the reason that SammyS gave, we have
f(8) - f(3) = 16
=> f(8) - 10 = 16 (It's given that f(3) = 10.)
=> f(8) = 26

 

[B18]

Thanks guys. I am going to sketch a picture or two so i fully understand FTC and how this problem was solved. Thanks again.

 

[Mark44]

I'm not sure a picture would be helpful, but don't let that stop you. The idea is pretty simple.

The 2nd part of the FTC is usually presented like this (with some of the fine print about continuity omitted)If F is an antiderivative of f, then
$$ \int_a^b f(x)dx = F(b) - F(a)$$

In your problem, f is an antiderivative of f', so
$$ \int_3^8 f'(x)dx = f(8) - f(3)$$

Antidifferentiation and differentiation are pretty much inverse operations, so if you antidifferentiate something that is already a derivative, you get back the original function.

 

[B18]

Your right a picture is not extremely beneficial. So in the definition of the FTC you provided the f(x)dx... the f(x) would be considered (a) the function and F(a) and F(b) are antiderivatives/integrals that are being evaluated at 3, 8 respectively. So to make sure i completely understand this for our test tomorrow, if we had ∫(3 to 8) f''(x) is it safe to assume that equals f'(8)-f'(3)

 

[B18]

Having trouble finding how to place the bounds on the integral correctly. Sorry :(

 

[Mark44]

Your right a picture is not extremely beneficial. So in the definition of the FTC you provided the f(x)dx... the f(x) would be considered (a) the function and F(a) and F(b) are antiderivatives/integrals that are being evaluated at 3, 8 respectively. So to make sure i completely understand this for our test tomorrow, if we had ∫(3 to 8) f''(x) is it safe to assume that equals f'(8)-f'(3)

Yes.

 

[Mark44]

Having trouble finding how to place the bounds on the integral correctly. Sorry :(

Sammy and I used LaTeX, which is essential for doing nice definite integrals. The script looks like this

[ tex]\int_{3}^{8} f(x) dx [/ tex]

To make this display without rendering as an integral, I had to insert a couple of extra spaces at the beginning of the tex and /tex tags. If you remove the spaces, it looks like this:
$$\int_{3}^{8} f(x) dx$$

Instead of [ tex ] and [ /tex ] tags, I usually use $ $ at the beginning and end, again without the space.