Question #dc81f

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Question #dc81f

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Answer 1 · 2017-11-22T06:15:05

$5 (\frac{1}{2 e^{2}} + 7)$ $5(1/(2e^2)+7)$ or $~ 35.33834$ $~35.33834$

Explanation:

let's make the exponential decay graph $f (x)$ $f(x)$ graph{e^(-0.4x) [-7.024, 7.024, -3.51, 3.514]}
and the linear graph $g (x)$ $g(x)$
graph{2.4x+1 [-7.024, 7.024, -3.51, 3.514]}

Judging by the graphs, the linear graph will be the "upper curve" from $0$ $0$ to $5$ $5$ since it is greater than the exponential decay graph at all points $(0, 5]$ $(0,5]$

So it will be integration of upper area minus lower area for the domain, or $\int_{0}^{5} g (x) - f (x) d x$ $int_0^5g(x)-f(x)dx$ .
Then you substitute the equations, so
$\int_{0}^{5} (2.4 x + 1) - (e^{- 0.4 x}) d x$ $int_0^5(2.4x+1)-(e^(-0.4x))dx$
= $\int_{0}^{5} 2.4 x + 1 - e^{- 0.4 x} d x$ $int_0^5 2.4x+1-e^(-0.4x)dx$

Then you can integrate and solve using the 2nd Fundamental Theorem of Calculus.

$\int_{0}^{5} 2.4 x + 1 - e^{- 0.4 x} d x$ $int_0^5 2.4x+1-e^(-0.4x)dx$
= $(1.2 x^{2} + x + 2.5 e^{- 0.4 x}) {∣ ∣}_{0}^{5}$ $(1.2x^2+x+2.5e^(-0.4x))|_0^5$
= $1.2 \cdot 5^{2} + 5 + 2.5 e^{- 0.4 \cdot 5} - (0)$ $1.2*5^2+5+2.5e^(-0.4*5)-(0)$
= $5 (\frac{1}{2 e^{2}} + 7)$ $5(1/(2e^2)+7)$ or $~ 35.33834$ $~35.33834$

Ask if you need any clarification on this. Thanks for asking though.

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