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
Evaluate the definite integral
4
x(x+3)dx
0
⌠(x^2 + 3x)dx from 0 to 4
⌡
= (x^3)/3 + (3/2)x from 0 to 4
= 64/3 + 6 - 0
= 82/3
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.