I am having a lot of trouble doing these types of problems. Can someone show me how to do this. Thank you for your time.

Evaluate the definite integral

4
x(x+3)dx
0

expand it first to get

⌠ (x^2 + 3x)dx from 0 to 4


= (x^3)/3 + (3/2)x from 0 to 4
= 64/3 + 6 - 0
= 82/3

Sure, I can help you with that! To evaluate the definite integral of the function x(x+3) with respect to x from 0 to 4, you can follow these steps:

Step 1: Simplify the integrand:
The integrand is x(x+3), so you can expand the expression by multiplying:
x(x+3) = x^2 + 3x

Step 2: Now you have simplified the integrand, you can rewrite the definite integral:

∫(x^2 + 3x) dx from 0 to 4

Step 3: Apply the power rule for integration:

The power rule states that the integral of x^n with respect to x is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.

For the term x^2, the integral would be:
(1/3) * x^3

For the term 3x, the integral would be:
(3/2) * x^2

Step 4: Evaluate the definite integrals:

To evaluate the definite integral from 0 to 4, you plug in the upper limit (4) and lower limit (0) into the antiderivative expression and subtract the results.

At x = 4:
(1/3) * 4^3 + (3/2) * 4^2

At x = 0:
(1/3) * 0^3 + (3/2) * 0^2

Step 5: Subtract the results:
Subtract the value obtained at the lower limit (x = 0) from the value obtained at the upper limit (x = 4).

So:

[(1/3) * 4^3 + (3/2) * 4^2] - [(1/3) * 0^3 + (3/2) * 0^2]

Simplifying this would give you the final answer.

I hope this helps! Let me know if you have any further questions.