MHB Evaluate Integral: Get Help Now!

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The integral to evaluate is ∞ ∫0.2e^−0.2u du. A substitution is made with t = 0.2u, leading to dt = 0.2 du and adjusted limits of integration from 2 to b. The evaluation involves taking the limit as b approaches infinity, resulting in the expression -[-e^{-b} + e^{-2}]. As b approaches infinity, the term e^{-b} approaches zero, simplifying the result to 1/e^2. The final answer for the integral is 1/e^2.
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Can someone help me with this? Not sure where to start.

Exercise 1 (integration) Evaluate the integral

∫0.2e^−0.2u du.
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$t = 0.2u \implies dt = 0.2 \, du$

substitute and reset the limits of integration

$\displaystyle \lim_{b \to \infty} \int_{2}^b e^{-t} \, dt$

$\displaystyle \lim_{b \to \infty} \bigg[-e^{-t} \bigg]_2^b$

$\displaystyle -\lim_{b \to \infty} \bigg[e^{-b} - e^{-2} \bigg] = -\bigg[0 - \dfrac{1}{e^2} \bigg] = \dfrac{1}{e^2}$
 

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