What is the integral from 1 to 4 of (5x^3-6) ?

The integral from 1 to 4 of (5x^3-6) is 1203/4

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integral from 1 to 4 of (5x^3-6)

\int_{1}^{4} (5 x^{3} - 6) d x

\frac{1203}{4}

Decimal Notation

300.75

Solution steps

\int_{1}^{4} 5 x^{3} - 6 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} 5 x^{3} d x - \int_{1}^{4} 6 d x

\int_{1}^{4} 5 x^{3} d x = \frac{1275}{4}

\int_{1}^{4} 6 d x = 18

= \frac{1275}{4} - 18

Simplify

\frac{1275}{4} - 18 : \frac{1203}{4}

= \frac{1203}{4}

\frac{d y}{d x} = 5 y e^{x + 8}

\int sec^{3} (7 x) d x

y^{^{'}} = sec (2) (x - y)

f (1) = 2 x^{4} - x^{3} + 6 x^{2}

x \to \infty lim (3 - x)

What is the integral from 1 to 4 of (5x^3-6) ?
The integral from 1 to 4 of (5x^3-6) is 1203/4