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

The integral from 2 to x of (3t^2-1) is x^3-x-6

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

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

x^{3} - x - 6

Solution steps

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

Apply the Sum Rule:

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

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

\int_{2}^{x} 3 t^{2} d t = x^{3} - 8

\int_{2}^{x} 1 d t = x - 2

= x^{3} - 8 - (x - 2)

Simplify

= x^{3} - x - 6

F (r) = \frac{5}{r ^{3}}

\frac{d}{d x} (\frac{3}{x + 7})

\int \frac{csc ^{4} ( 2 x )}{cot ^{2} ( 2 x )} d x

x \frac{d y}{d x} = y + (x^{2} - 1)^{2}

What is the integral from 2 to x of (3t^2-1) ?
The integral from 2 to x of (3t^2-1) is x^3-x-6