Question

evaluate the integral.
int_{0}^{4}(6x - 3)dx

Explanation:

Step1: Find antiderivative

The antiderivative of \(6x - 3\) is \(F(x)=6\times\frac{x^{2}}{2}-3x = 3x^{2}-3x\) using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).

Step2: Apply fundamental theorem of calculus

\(\int_{0}^{4}(6x - 3)dx=F(4)-F(0)\).
Substitute \(x = 4\) into \(F(x)\): \(F(4)=3\times4^{2}-3\times4=3\times16 - 12=48-12 = 36\).
Substitute \(x = 0\) into \(F(x)\): \(F(0)=3\times0^{2}-3\times0 = 0\).
Then \(F(4)-F(0)=36-0 = 36\).

Answer:

36