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.
10
 

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$t = 0.2u \implies dt = 0.2 \, du$

substitute and reset the limits of integration

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

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

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

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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