Question

Evaluate the definite integral


1. Integral of dx / x^2

Upper: 2

Lower; -2


2. Integral of 2^x dx

upper; 0

lower: Negative infinity

Answer 1

1. Evaluate the definite integral

$\int_{-2}^2 \frac{dx}{x^2} = 2 \int_{0}^2 \frac{dx}{x^2} = 2 \lim_{\epsilon \to 0+} \int_{\epsilon}^2 \frac{dx}{x^2} \\= 2 \lim_{\epsilon \to 0+} \left(- \frac{1}{x}\right)\vert_{\epsilon}^2 = +\infty$

The first equality holds because the function $\frac{1}{x^2} ,\; x \in [-2;2]$ is even.

The rest follows from improper integral definition and the fundamental theorem of calculus.

Answer: $+\infty$ .

2. Evaluate the definite integral

$\int_{-\infty}^0 2^x dx = \int_{-\infty}^0 e^{\ln(2)x} dx \\= \lim_{\Delta \to -\infty} \int_{\Delta}^0 e^{\ln(2)x} dx \\= \lim_{\Delta \to -\infty} \left(\frac{ e^{\ln(2)x}}{\ln(2)}\right) \vert_{\Delta}^0 = \frac{1}{\ln(2)}$

Answer: $\frac{1}{ \ln(2)}$