Answer to Question #100015 in Calculus for yenco

Question #100015
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
1
Expert's answer
2019-12-09T10:34:18-0500


  1. Evaluate the definite integral

22dxx2=202dxx2=2limϵ0+ϵ2dxx2=2limϵ0+(1x)ϵ2=+\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 1x2,  x[2;2]\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

02xdx=0eln(2)xdx=limΔΔ0eln(2)xdx=limΔ(eln(2)xln(2))Δ0=1ln(2)\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: 1ln(2)\frac{1}{ \ln(2)}


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