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

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

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

- \frac{2}{\sqrt{e^{3}}} + 2 \sqrt{e}

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

Question 2 of 2

Find the integral

\int_{1}^{2} 5 e^{2 x - 1} d x

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

1. $\frac{5}{2} [e^{2} - 1]$ $5/2 [e^2-1]$
2. $5 e [e^{2} - 1]$ $5e[e^2-1]$
3. $e^{2} - 1$ $e^2-1$
4. $\frac{5}{2} e [e^{2} - 1]$ $5/2 e[e^2-1]$

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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

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

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

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

x

$x$ .

	${[\frac{5 e^{2 x - 1}}{2}]}_{1}^{2}$ $[(5e^(2x-1))/2]_1^2$

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

$=$ $=$	$\frac{5 e^{3}}{2} - \frac{5 e}{2}$ $(5e^3)/2-(5e)/2$	Evaluate

$=$ $=$	$\frac{5}{2} e [e^{2} - 1]$ $5/2 e[e^2-1]$	Factorise

\frac{5}{2} e [e^{2} - 1]

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

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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