Question

can someone show me how to get the answer 1/2 on this question

Answer 1

\[\int\limits_{0}^{\infty} e^{-2x} dx\]

Answer 2

First, to solve the indefinite integral, \[\int e^{-2x}dx\] we let \[u=-2x\] so that du=-2dx, or dx=-1/2 du, so that: \[\int e^{-2x}dx = -\frac{1}{2}\int e^u du = -\frac{1}{2}e^u+C\] Now, evaluating the definite integral, note that when x=0, u=0, and when x=∞, then u=-∞, so: \[\int_0^\infty e^{-2x}dx=-\frac{1}{2}\int_0^{-\infty} e^u du\] and this is equal to \[ -\frac{1}{2}(e^{-\infty}-e^0) = -\frac{1}{2}(0-1) = \frac{1}{2}\]