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Intergration

Evaluate:

Unparseable latex formula:

\int^5_1(2x-1)^3\dx

Heres what i did.

[\frac{1}{4}(2x-1)^4]^5_1

\frac{1}{4}(9)^4-\frac{1}{4}(1)^4=1640.25

The book gets half of that answer: 820

Reply 1

Gemini92

Original post by Emissionspectra

Evaluate:

Unparseable latex formula:

\int^5_1(2x-1)^3\dx

Heres what i did.

[\frac{1}{4}(2x-1)^4]^5_1

\frac{1}{4}(9)^4-\frac{1}{4}(1)^4=1640.25

The book gets half of that answer: 820

You haven't integrated correctly. Try using the substitution u=(2x-1)

Reply 2

jameswhughes

Think of the chain rule, backwards. When you differentiate you'll get an extra 2 from the bracket.

Reply 3

Emissionspectra

Original post by Gemini92

You haven't integrated correctly. Try using the substitution u=(2x-1)

Ah ok thanks :smile:

, on a side note can you explain this I dont see how

\frac{d}{dx}e^2x

Is of the form

\frac{d}{dx}f(ax+b)

as the book says it is...

Reply 4

ElMoro

Original post by Emissionspectra

Ah ok thanks :smile:

, on a side note can you explain this I dont see how

\frac{d}{dx}e^2x

Is of the form

\frac{d}{dx}f(ax+b)

as the book says it is...

I'm assuming you mean

\frac{d}{dx} \left(e^{2x}\right)

f(x) could be any function in this case lets say

f(x) = e^x

That means that

f(ax + b) = e^{ax+b}

Does that help? :smile:

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