I am having trouble doing this problem. I know how to do indefinate integrals, but I don't know how to do definate integrals. Can you show me how to do this.

Evaluate
5
(x^3-2x)dx
2

To get the definite integral subtract the value of the indefinite integral at x=2 from the value at x=5. The arbitrary cnstant c will cancel out.

The indefinite integral is x^4/4 - x^2 + C.

The definite integral is
(625/4) - 25 -[16/4 - 4].

To evaluate a definite integral, you need to use the fundamental theorem of calculus. This theorem states that if a function, f(x), is continuous on the interval [a, b] and F(x) is an antiderivative of f(x), then:

∫[a to b] f(x)dx = F(b) - F(a)

In other words, to evaluate the definite integral, you need to find an antiderivative of the integrand function and then evaluate it at the upper limit (b) and subtract the value of the antiderivative at the lower limit (a).

In this case, you need to evaluate the definite integral of the function f(x) = x^3 - 2x from x = 2 to x = 5.

First, find the antiderivative of f(x). For each term in the integrand, apply the power rule of integration:

∫ x^n dx = (1/(n+1)) * x^(n+1)

Using this rule, the antiderivative of x^3 is (1/4) * x^4, and the antiderivative of -2x is -x^2.

Now, evaluate the antiderivative at x = 5 and subtract the value at x = 2:

F(5) - F(2) = (1/4)(5^4) - (-5^2) - [(1/4)(2^4) - (-2^2)]
= (1/4)(625) - 25 - (1/4)(16) - 4
= (625/4) - 25 - 4 - 4
= 625/4 - 25 - 4 - 4
= 155.25 - 33
= 122.25

Therefore, the value of the definite integral of (x^3 - 2x)dx from x = 2 to x = 5 is 122.25.