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

The integral from 1 to 4 of (x^3-3x^2+1) is 15/4

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

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

\frac{15}{4}

Solution steps

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

\int_{1}^{4} x^{3} d x = \frac{255}{4}

\int_{1}^{4} 3 x^{2} d x = 63

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

= \frac{255}{4} - 63 + 3

Simplify

\frac{255}{4} - 63 + 3 : \frac{15}{4}

= \frac{15}{4}

\int_{0}^{- 1} x \cdot e^{x} d x

\int_{0}^{2} (x + 2)^{2} - 3 d x

\int_{5}^{10} 1

\int_{- 2}^{0} \frac{- 2 x}{1 + x ^{2}} d x

\int_{0}^{\frac{t}{3}} e^{t^{2}} d t

What is the integral from 1 to 4 of (x^3-3x^2+1) ?
The integral from 1 to 4 of (x^3-3x^2+1) is 15/4