Question

22 multiple choice 1 point evaluate. \\( \int_{1}^{e} \left( 20x - \frac{9}{x} \
ight) dx \\) \\( 10e^2 - 19 \\) \\( 10e^2 - 10 \\) \\( 20e^2 - 9 \\) \\( 10e^2 - 9 \\)

Explanation:

Step1: Integrate term by term

The integral of \(20x\) with respect to \(x\) is \(10x^2\) (using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}\), here \(n = 1\), so \(\int20x dx=20\times\frac{x^{2}}{2}=10x^{2}\)), and the integral of \(-\frac{9}{x}\) with respect to \(x\) is \(- 9\ln|x|\) (since \(\int\frac{1}{x}dx=\ln|x|+C\)). So the antiderivative \(F(x)\) of \(20x-\frac{9}{x}\) is \(10x^{2}-9\ln x\).

Step2: Apply Fundamental Theorem of Calculus

We evaluate \(F(e)-F(1)\). First, \(F(e)=10e^{2}-9\ln e=10e^{2}-9\) (since \(\ln e = 1\)). Then, \(F(1)=10(1)^{2}-9\ln1=10 - 0 = 10\) (since \(\ln1 = 0\)). Now, \(F(e)-F(1)=(10e^{2}-9)-10=10e^{2}-19\).

Answer:

\(10e^{2}-19\) (the first option among the given choices)