Question

evaluate the definite integral. use a graphing utility to verify your result.
$$ int _ { - 1 } ^ { 0 } ( 2 x - 3 ) d x $$

Explanation:

Step1: Find the antiderivative

The antiderivative of \(2x\) is \(x^{2}\) (using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}\), for \(n = 1\), \(\int2x dx=2\times\frac{x^{2}}{2}=x^{2}\)) and the antiderivative of \(- 3\) is \(-3x\) (since \(\int kdx=kx\) for constant \(k\)). So the antiderivative of \(2x - 3\) is \(F(x)=x^{2}-3x\).

Step2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus for definite integrals states that \(\int_{a}^{b}f(x)dx=F(b)-F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\). Here, \(a=- 1\), \(b = 0\) and \(f(x)=2x - 3\), \(F(x)=x^{2}-3x\).

First, calculate \(F(0)\):
\(F(0)=0^{2}-3\times0 = 0\)

Then, calculate \(F(-1)\):
\(F(-1)=(-1)^{2}-3\times(-1)=1 + 3=4\)

Now, \(\int_{-1}^{0}(2x - 3)dx=F(0)-F(-1)=0 - 4=-4\)

Answer:

\(-4\)