Integrals - AP Calculus AB

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

Here we are using several basic properties of definite integrals as well as the fundamental theorem of calculus.

First, you can pull coefficients out to the front of integrals.

Second, we notice that our lower bound is bigger than our upper bound. You can switch the upper and lower bounds if you also switch the sign.

Lastly, our integral "distributes" over addition and subtraction. That means you can split the integral by each term and integrate each term separately.

Now we integrate and calculate using the Fundamental Theorem of Calculus.

We proceed as follows

. (Start)

. (Break up the integral using the additive rule.)

. (We don't have information about the 2nd integral, so we solve our first equation for and replace it in this integral.)

. (Factor out the using linearity.)

. (Substitute in what we were given.)

.