Integral of s(3x²-2x+1)dx dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫s((3x2−2x)+1)dx=s∫((3x2−2x)+1)dx
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Integrate term-by-term:
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫3x2dx=3∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: x3
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2x)dx=−2∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: −x2
The result is: x3−x2
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
The result is: x3−x2+x
So, the result is: s(x3−x2+x)
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Now simplify:
sx(x2−x+1)
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Add the constant of integration:
sx(x2−x+1)+constant
The answer is:
sx(x2−x+1)+constant
The answer (Indefinite)
[src]
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| s*\3*x - 2*x + 1/ dx = C + s*\x + x - x /
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∫s((3x2−2x)+1)dx=C+s(x3−x2+x)
Use the examples entering the upper and lower limits of integration.