What is the integral from 1 to 3 of (4x^3-8x+1) ?

The integral from 1 to 3 of (4x^3-8x+1) is 50

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integral from 1 to 3 of (4x^3-8x+1)

\int_{1}^{3} (4 x^{3} - 8 x + 1) d x

50

Solution steps

\int_{1}^{3} 4 x^{3} - 8 x + 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}^{3} 4 x^{3} d x - \int_{1}^{3} 8 x d x + \int_{1}^{3} 1 d x

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

\int_{1}^{3} 8 x d x = 32

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

= 80 - 32 + 2

Simplify

= 50

\frac{d}{d x} (\frac{4}{x} + x^{2})

\int \frac{1}{5 - 3 cos ( x )} d x

\int 8 e^{4 x} (e^{4 x} - 4)^{\frac{1}{5}} d x

y^{^{'}} + \frac{1}{x} y = 5 x^{2}

2 y^{^{''}} + 26 y = 0, y (0) = 9, y^{^{'}} (0) = 5

What is the integral from 1 to 3 of (4x^3-8x+1) ?
The integral from 1 to 3 of (4x^3-8x+1) is 50