Question

fundamental theorem of calculus: p
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evaluate the definite integral:
$\int_{-3}^{3}(6x - e^{x})dx = 0$
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Explanation:

Step1: Find the antiderivative

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

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Answer:

\(\frac{1}{e^{3}}-e^{3}\)