Using the Fundamental Theorem of Calculus (FTC), compute the following:

Question

Using the Fundamental Theorem of Calculus (FTC), compute the following:

1. integrate (x ^ 3 + 5) dx from 2 to 3

2. integrate (x ^ 2 + 1) dx from 2 to 6

Answer 1

The Fundamental Theorem of Calculus tells us that the value of the definite integral of f(x) with bounds a and b is F(b) - F(a) where F'(x) = f(x). In other words, F(x) is the antiderivative of f(x). So to solve these integrals, we need to find the antiderivative of the functions and plug in the bounds.

1. In this example, f(x) = x^3 + 5. In order to take the antiderivative of a term of a polynomial, you raise its exponent by 1 and divide the term by the new exponent. For the term x^3, we first raise its power to x^4 and then divide the whole thing by 4 to get (x^4)/4. For the term 5, which is the same as 5x^0, we raise the term once to 5x^1 and then divide by 1 to get 5x. So the total antiderivative F(x) = (x^4)/4 + 5x. Now just plug in the bounds to get F(3) - F(2) = ((3^4)/4 + 5*3) - ((2^4)/4 + 5*2) = (81/4 + 15) - (4 + 10) = 21.25.
2. Let's find the antiderivitaive again. For the x^2 term, raise the power to x^3 and divide by 3 to get (x^3)/3. For the 1x^0 term, raise the power to 1x^1 and divide by 1 to get x. So F(x) = (x^3)/3 + x. Let's plug in bounds to get F(6) - F(2) = ((6^3)/3 + 6) - ((2^3)/3) + 2) = 78 - 14/3 = 73 and 1/3.