- #1
bigevil
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Please help me check my solution. Is there a simpler way to do this?
Integrate x^3 cos x from first principles.
Taylor series of cos x
Sum of powers from 1 to n
I am dividing the area under the curve from a to b into n strips and then summing up the areas (where the area is x.f(x)). Then express this in terms of n and let n tend to infinity.
Let Sn = (Sigma) f(x') (delta)x
where
f(x) = x^3 cos x
f(x) = x^3 - x^5/2! + x^7/4!
x' = j . (delta)x
Insert values from 1 to n, and then group
...
...
Sn = [(delta)x]^4 . (1^3 + 2^3... + n^3)
+ [(delta)x]^6 . (1/2!) . (1^5 + 2^5 ... + n^5)
+ [(delta)x]^8 . (1/4!) . (1^7 + 2^7 ... + n^7)
Delta x = b / n, where b is the upper limit for integration
which yields 1/4 b^4 + 1/12 b^6...
Homework Statement
Integrate x^3 cos x from first principles.
Homework Equations
Taylor series of cos x
Sum of powers from 1 to n
The Attempt at a Solution
I am dividing the area under the curve from a to b into n strips and then summing up the areas (where the area is x.f(x)). Then express this in terms of n and let n tend to infinity.
Let Sn = (Sigma) f(x') (delta)x
where
f(x) = x^3 cos x
f(x) = x^3 - x^5/2! + x^7/4!
x' = j . (delta)x
Insert values from 1 to n, and then group
...
...
Sn = [(delta)x]^4 . (1^3 + 2^3... + n^3)
+ [(delta)x]^6 . (1/2!) . (1^5 + 2^5 ... + n^5)
+ [(delta)x]^8 . (1/4!) . (1^7 + 2^7 ... + n^7)
Delta x = b / n, where b is the upper limit for integration
which yields 1/4 b^4 + 1/12 b^6...