Integral Problem

[Kevrad]

Compute f'(1) where f(x) is defined by
f(x)=(integral Sign)sqrt[1+x^3]dx

For this problem do i just plug in 1 to sqrt[1+x^3]?

 

[daon2]

Compute f'(1) where f(x) is defined by
f(x)=(integral Sign)sqrt[1+x^3]dx

For this problem do i just plug in 1 to sqrt[1+x^3]?

Yes, the derivative is the left inverse to an antiderivative. So:

\(\displaystyle \displaystyle \dfrac{d}{dx}\left( \underbrace{\int\, g(x)\, dx}_{f(x)} \right) = \underbrace{g(x)}_{f'(x)}\)

 

[HallsofIvy]

Compute f'(1)

I can see how you would have difficulty with this!

 

[dsk2]

Compute f'(1) where f(x) is defined by
f(x)=(integral Sign)sqrt[1+x^3]dx

For this problem do i just plug in 1 to sqrt[1+x^3]?

Given: f(x)=integral of sqrt (1+x^3) dx

Thus, f'(x)=sqrt(1+x^3)
Hence, f'(1)=sqrt(1+1^3)=sqrt(2)=1.414

 

[lookagain]

Given: f(x) = integral of sqrt (1 + x^3) dx

Thus, f'(x) = sqrt(1 + x^3)

Hence, f'(1) = sqrt(1 + 1^3) = sqrt(2) \(\displaystyle \approx\) 1.414

The equals sign was traded for the "is approximately equal to" sign.

And spacing was added for more readability.