Integral of (x-3)^3 dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Let u=x−3.
Then let du=dx and substitute du:
∫u3du
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The integral of un is n+1un+1 when n=−1:
∫u3du=4u4
Now substitute u back in:
4(x−3)4
Method #2
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Rewrite the integrand:
(x−3)3=x3−9x2+27x−27
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Integrate term-by-term:
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The integral of xn is n+1xn+1 when n=−1:
∫x3dx=4x4
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The integral of a constant times a function is the constant times the integral of the function:
∫(−9x2)dx=−9∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: −3x3
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The integral of a constant times a function is the constant times the integral of the function:
∫27xdx=27∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: 227x2
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The integral of a constant is the constant times the variable of integration:
∫(−27)dx=−27x
The result is: 4x4−3x3+227x2−27x
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Now simplify:
4(x−3)4
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Add the constant of integration:
4(x−3)4+constant
The answer is:
4(x−3)4+constant
The answer (Indefinite)
[src]
/
| 4
| 3 (x - 3)
| (x - 3) dx = C + --------
| 4
/
∫(x−3)3dx=C+4(x−3)4
The graph
Use the examples entering the upper and lower limits of integration.