What is the integral from-1 to 5 of (4x^3-3x^2+1) ?

The integral from-1 to 5 of (4x^3-3x^2+1) is 504

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integral from-1 to 5 of (4x^3-3x^2+1)

\int_{- 1}^{5} (4 x^{3} - 3 x^{2} + 1) d x

504

Solution steps

\int_{- 1}^{5} 4 x^{3} - 3 x^{2} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{5} 4 x^{3} d x - \int_{- 1}^{5} 3 x^{2} d x + \int_{- 1}^{5} 1 d x

\int_{- 1}^{5} 4 x^{3} d x = 624

\int_{- 1}^{5} 3 x^{2} d x = 126

\int_{- 1}^{5} 1 d x = 6

= 624 - 126 + 6

Simplify

= 504

\int_{0}^{2} x e^{\frac{x ^{2}}{2}} d x

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\int_{- 1}^{1} (3 + 2 x)^{2} d x

\int_{0}^{2} \frac{x ^{3}}{3} d x

\int_{1}^{e^{3}} \frac{3}{x} d x

What is the integral from-1 to 5 of (4x^3-3x^2+1) ?
The integral from-1 to 5 of (4x^3-3x^2+1) is 504