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

The integral from 1 to 3 of (4x^3-3x^2) is 54

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

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

54

Solution steps

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

\int_{1}^{3} 4 x^{3} d x = 80

\int_{1}^{3} 3 x^{2} d x = 26

= 80 - 26

Simplify

= 54

\frac{1}{s ^{2} + s}

\frac{d ^{2} y}{d t ^{2}} - e^{2} y = 0

\int \frac{u}{u ^{2} + 2} d u

\int_{0}^{1} x + 4 d x

\int (3 x + 4) e^{- x} d x

What is the integral from 1 to 3 of (4x^3-3x^2) ?
The integral from 1 to 3 of (4x^3-3x^2) is 54