Question

g(x) = integral(1 to x) 1/(t^3 + 1)dt

Answer 1

Use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

Answer 2

WAIT NO I FORGOT THE DERIVATIVE
\[\frac{d}{dx}\left [ \int_k^t f(x) dx \right ] = f(t) · f'(t)\]

Answer 3

Note that this only works if the bottom number is a constant (1 in this case). If is on the bottom and the constant is on top, then flip and negate the entire integral before you do anything.

If both bounds are variables, then the equation changes:

\[\frac{d}{dx}\left [ \int_{t(x)}^{u(x)} f(x) dx \right ] = f(u(x)) · f'(u(x)) - f(t(x)) · f'(t(x))\]

Answer 4

Sorry, I'm still a little confused on what is going on. Could you give me a step-to-step scenario?

Answer 5

Okay so there are three scenarios:

1) The top bound (the little tiny number near the top of the integral) is a variable and the bottom bound is a number, like d/dx[integral (1 to 5x) t dt ]. If so:
\[\frac{d}{dx} \left [ \int_1^{5x} t^2 \, dt\right ] = (5x)^2 ·5\]
Basically what you do is plug the top function into the \(t^2\) function and multiply that to what you get when you plug the top function into the derivative of the top bound.

2) The second scenario is when the top bound is a constant and the bottom bound is a function:

\[\frac{d}{dx} \left [ \int_{x^2}^1 3t \,\, dt\right ] = \frac{d}{dx} \left [ - \int_1^{x^2} 3t \, dt\right ] = -3x^2 ·2x\]

3) The third scenario is when both bounds are functions:

\[\frac{d}{dx} \left [ \int_{x^2}^{x^3} t \,\, dt\right ] = x^3·3x^2 - x^2·2x\]

4) (Trick question) What happens if both bounds are numbers?

Answer 6

FYI, for #1 and #2, it doesn't matter WHAT the constant is, just that it IS a constant.

Answer 7

Oh, I get it now. Thanks so much, Kevin!