Question

evaluate the definite integral. \\(\int_{-1}^{1} (9x^2 + 6x - 3) dx = \\)

Explanation:

Step1: Find the antiderivative

The antiderivative of \(9x^2\) is \(9\times\frac{x^3}{3}=3x^3\), the antiderivative of \(6x\) is \(6\times\frac{x^2}{2}=3x^2\), and the antiderivative of \(-3\) is \(-3x\). So the antiderivative \(F(x)\) of \(f(x)=9x^2 + 6x - 3\) is \(F(x)=3x^3+3x^2 - 3x\).

Step2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that \(\int_{a}^{b}f(x)dx=F(b)-F(a)\). Here, \(a = - 1\), \(b = 1\), so we calculate \(F(1)-F(-1)\).

First, calculate \(F(1)\):
\(F(1)=3(1)^3+3(1)^2-3(1)=3 + 3-3=3\)

Then, calculate \(F(-1)\):
\(F(-1)=3(-1)^3+3(-1)^2-3(-1)=-3 + 3 + 3=3\)

Now, find \(F(1)-F(-1)\):
\(F(1)-F(-1)=3 - 3=0\)

Answer:

\(0\)