Question

fundamental theorem of calculus
(1 point)
evaluate the definite integral:
$$ int _ { 3 } ^ { 3 } ( 6 x - e ^ { x } ) d x = $$

Explanation:

Step1: Find antiderivative

The antiderivative of \(6x\) is \(3x^{2}\) (using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C,n
eq - 1\), here \(n = 1\)), and the antiderivative of \(e^{x}\) is \(e^{x}\). So the antiderivative of \(6x-e^{x}\) is \(F(x)=3x^{2}-e^{x}\).

Step2: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus \(\int_{a}^{b}f(x)dx=F(b)-F(a)\). Here \(a = 3\), \(b = 3\). Then \(\int_{3}^{3}(6x - e^{x})dx=(3x^{2}-e^{x})\big|_{3}^{3}\).
Substitute \(x = 3\) into \(F(x)\): \(F(3)=3\times3^{2}-e^{3}=27 - e^{3}\), substitute \(x = 3\) again (since upper and lower limits are the same). Then \(F(3)-F(3)=(27 - e^{3})-(27 - e^{3})\).

Answer:

\(0\)