Evaluate the definite integral ∫ from 0 to 2 of (3x^2 + 1) dx.
Solution
Problem
Evaluate the definite integral ∫ from 0 to 2 of (3x^2 + 1) dx.
Step 1: Find an antiderivative.
We need to evaluate:
∫ from 0 to 2 of (3x^2 + 1) dx
Integrate each term:
∫3x^2 dx = x^3
and:
∫1 dx = x
So an antiderivative is:
F(x) = x^3 + x
Step 2: Apply the Fundamental Theorem of Calculus.
For a definite integral:
∫ from a to b of f(x) dx = F(b) - F(a)
Here, a = 0 and b = 2.
Therefore:
F(2) - F(0)
Step 3: Evaluate the upper endpoint.
F(2) = 2^3 + 2
F(2) = 8 + 2
F(2) = 10
Step 4: Evaluate the lower endpoint.
F(0) = 0^3 + 0
F(0) = 0
Step 5: Subtract.
10 - 0 = 10
Therefore, the value of the definite integral is 10.
Result
10