Definite Integrals of Exponential Functions

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Question 1 of 2

Find the integral

\int_{- 1}^{3} e^{- \frac{x}{2}} d x

$\int_{-1}^{3} e^{-\frac{x}{2}} dx$

1. $- \frac{2}{\sqrt{e^{3}}} + 2 \sqrt{e}$ $-2/(sqrt(e^3)) +2sqrte$
2. $- \frac{2}{\sqrt{e^{3}}} + 2 e$ $-2/(sqrt(e^3)) +2e$
3. $e^{2} - 1$ $e^2-1$
4. $\frac{2}{\sqrt{e^{3}}} + 2 \sqrt{e}$ $2/(sqrt(e^3)) +2sqrte$

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Integrating Exponential Functions with Base “e”

\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c

$\int e^{\color{#004ec4}{a}x+b} dx=\frac{1}{\color{#004ec4}{a}} e^{\color{#004ec4}{a}x+b} +c$

Substitute the components into the formula

$\int e^{\color{#004ec4}{a}x+b} dx$	$=$ $=$	$\frac{1}{\color{#004ec4}{a}} e^{\color{#004ec4}{a}x+b} +c$

$\int_{-1}^{3} e^{\color{#004ec4}{-\frac{x}{2}}} dx$	$=$ $=$	$\int_{-1}^{3} \frac{e^{\color{#004ec4}{-\frac{x}{2}}}}{-\frac{1}{2}}$	Substitute known values

	$=$ $=$	${[- 2 e^{- \frac{x}{2}}]}_{- 1}^{- \frac{x}{2}}$ $[-2e^(-x/2)]_(-1)^3$	Simplify

Finally, get the difference of the upper and lower limits substituted to the integral as

x

$x$ .

	${[- 2 e^{- \frac{x}{2}}]}_{- 1}^{- \frac{x}{2}}$ $[-2e^(-x/2)]_(-1)^3$

$=$ $=$	$[- 2 e^{- \frac{3}{2}}] - [- 2 e^{- \frac{- 1}{2}}]$ $[-2e^(-3/2)]-[-2e^(- (-1)/2)]$	Substitute the limits

$=$ $=$	$- 2 e^{- \frac{3}{2}} + 2 e^{\frac{1}{2}}$ $-2e^(-3/2)+2e^(1/2)$	Evaluate

$=$	$-2/e^(3/2) +2e^(1/2)$	Reciprocate $e^(-3/2)$

$=$	$-2/(sqrt(e^3)) +2sqrte$	Change the exponents into surds

$-2/(sqrt(e^3)) +2sqrte$

Question 2 of 2

Find the integral

$\int_{1}^{2} 5e^{2x-1} dx$

1. $5e[e^2-1]$
2. $e^2-1$
3. $5/2 e[e^2-1]$
4. $5/2 [e^2-1]$

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Integrating Exponential Functions with Base “e”

$\int e^{\color{#004ec4}{a}x+b} dx=\frac{1}{\color{#004ec4}{a}} e^{\color{#004ec4}{a}x+b} +c$

Substitute the components into the formula

$\int e^{\color{#004ec4}{a}x+b} dx$	$=$	$\frac{1}{\color{#004ec4}{a}} e^{\color{#004ec4}{a}x+b} +c$

$\int_{1}^{2} 5e^{\color{#004ec4}{2}x-1} dx$	$=$	$\left[\frac{5e^{\color{#004ec4}{2}x-1}}{2}\right]_{1}^{2}$	Substitute known values

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$[(5e^(2x-1))/2]_1^2$

$=$	$[(5e^(2(2)-1))/2]-[(5e^(2(1)-1))/2]$	Substitute the limits

$=$	$(5e^3)/2-(5e)/2$	Evaluate

$=$	$5/2 e[e^2-1]$	Factorise

$5/2 e[e^2-1]$

Definite Integrals of Exponential Functions

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Quiz summary

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1. Question

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Well Done!

Integrating Exponential Functions with Base “e”

2. Question

Hint

Well Done!

Integrating Exponential Functions with Base “e”

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