∫(1/2 - 3x +x^2)dx can anyone show me how to solve this.. Thank you
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∫1/2dx - ∫3xdx + ∫x^2dx
1/2 ∫dx - 3∫xdx + ∫x^2dx
1/2(x) - 3(x^2/2) + (x^3/3) + C
or x/2- 3/2(x^2) + x^3/3 + C
1/2 ∫dx - 3∫xdx + ∫x^2dx
1/2(x) - 3(x^2/2) + (x^3/3) + C
or x/2- 3/2(x^2) + x^3/3 + C
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well break this in 3 parts
∫(1/2)dx - ∫(3x)dx + ∫(x^2)dx
take constants outside and apply the formula ∫(x^n)dx = x^(n+1) / (n+1) + C
so you will get
(1/2)*x - 3*(x^2)/2 + (x^3)/3 + C
∫(1/2)dx - ∫(3x)dx + ∫(x^2)dx
take constants outside and apply the formula ∫(x^n)dx = x^(n+1) / (n+1) + C
so you will get
(1/2)*x - 3*(x^2)/2 + (x^3)/3 + C
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∫(1/2 - 3x +x^2)dx
=1/2x - 3/2x^2 + 1/3x^3 + C answer//
=1/2x - 3/2x^2 + 1/3x^3 + C answer//
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Use the power rule for derivatives in reverse.
∫(1/2 - 3x +x^2)dx =
x/2 - 3x^2/2 + x^3/3 + c
∫(1/2 - 3x +x^2)dx =
x/2 - 3x^2/2 + x^3/3 + c
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∫(1/2 - 3x +x²)dx = 1/2∫dx - 3∫x dx + ∫x² dx
= 1/2 x - 3/2 x² + x³/3 + C
= 1/2 x - 3/2 x² + x³/3 + C