Mathematics

# Evaluate the following integral:$\int { { 3 }^{ x } } dx$

##### SOLUTION
Given $\displaystyle \int{{3}^{x}dx}$

Let $t={3}^{x}$

$\Rightarrow\,\log{t}=x\log{3}$

$\Rightarrow\,\dfrac{dt}{t}=\log{3}dx$

$\Rightarrow\,\dfrac{1}{\log{3}}dt=tdx$

$\Rightarrow\,\dfrac{1}{\log{3}}dt={3}^{x}dx$

$\displaystyle \int{{3}^{x}dx}$
$=\displaystyle \int{\dfrac{1}{\log{3}}dt}$

$=\dfrac{1}{\log{3}}t+c$ where $c$ is the constant of integration
$=\dfrac{1}{\log{3}}{3}^{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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$\displaystyle \overset{\pi/2}{\underset{0}{\int}} \dfrac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ equals-
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1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate the given integral.
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If $\displaystyle \int _0^{\pi/2} \sin x \cos x dx$ is equal to:
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Q4 One Word Medium
$\displaystyle \int_{0}^{\pi /2}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx= \left ( a+b \right )\frac{\pi }{C}$
What is C?

Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$