Integrals of Exponential Functions
Key Questions
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By rewriting a bit,
#int n^x dx=int e^{(lnn)x}dx=e^{(lnn)x}/{lnn}+C=n^x/{lnn}+C# I hope that this was helpful.
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The answer is
#I=-e^(-x)+C# .This integral can be solved by a substitution:
#u=-x#
#du=-dx#
#-du=dx# So, now we can substitute:
#int e^(-x)dx = int e^u (-du)#
#=-int e^u du#
#=-e^u + C# and substitute back:
#=-e^(-x) + C# For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.
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Answer:
#int3^xdx=1/ln3 3^x+"c"# Explanation:
We want to find
#int3^xdx# .Make the natural substitution
#u=3^x# so#du=3^xln3dx# .So
#int3^xdx=1/ln3int1du=1/ln3 u+"c"=1/ln3 3^x+"c"# -
Since
#(e^x)'=e^x# ,we have
#int e^x dx=e^x+C# .I hope that this was helpful.
Questions
Introduction to Integration
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Sigma Notation
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Integration: the Area Problem
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Formal Definition of the Definite Integral
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Definite and indefinite integrals
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Integrals of Polynomial functions
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Determining Basic Rates of Change Using Integrals
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Integrals of Trigonometric Functions
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Integrals of Exponential Functions
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Integrals of Rational Functions
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The Fundamental Theorem of Calculus
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Basic Properties of Definite Integrals