 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Fundamental Theorem of Calculus - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The fundamental theorem of calculus states that if a function f has an antiderivative F, then the definite integral of f from a to b is equal to F(b)-F(a).

Find the antiderivative, and take the integral of the function using the methods of integration such as the power rule, knowledge of specific integrals, or substitution. Then, evaluate at the limits of integration. Remember that when taking definite integrals, it is possible for the result to be negative (this means that the curve is below the x-axis).

Let’s solve some more problems using the fundamental theorem of calculus. Let’s recall that the fundamental theorem requires what the antiderivative of your function is. So the definite integral from a to b of your function, is that antiderivative evaluated at b minus the antiderivative of the value evaluated at a, the lower limit. Let’s try it out.

Here is my first problem; integrate from 1 to 3, x over 4 plus 4 over x. I need to identify an antiderivative of this function. Actually, before I do this, let me rewrite the integral from 1 to 3. This is the same as 1/4x plus 4 times 1/x which is x to the -1. Now I can identify antiderivatives.

Here my antiderivative is going to be ¼ times ½ x², so 1/8x² plus 4 times, and the antiderivative for x to the -1, remember that’s the special power of x. it’s antiderivative is the natural log of the absolute value of x, from 1 to 3.

So first, 3; I plug in 3 and I get 1/8, 3² is 9 plus 4 times the natural log of 3, minus, and now I plug in 1. 1/8 times 1, which is 1/8 plus, 4 times natural log 1. Natural log 1 is zero. So 4 times 0, 0. And here I’ve got 1/8 times 9 minus 1/8, that’s 8/8 or 1. So my answer is 1 plus 4ln3. That’s the exact value of this definite integral.

Let’s take a look at another example. The integral from 0 to 1 of e to the x minus x. So this one is ready to go. All I need is to find an antiderivative. The antiderivative of e to the x is e to the x. The antiderivative for x is 1/2x². I evaluate that from 0 to 1.

First 1; e to the 1 is e minus ½ of 1², so minus ½ minus e to the 0 is 1 minus 0. So e minus ½ minus 1, that’s e minus 3/2. That’s it. Fundamental theory of calculus. Find an antiderivative for your function and evaluate that antiderivative at the two limits. First the upper limit, then the lower limit, and that’s your answer.