What is the integral from 1 to 4 of (5-2t+3t^2) ?

The integral from 1 to 4 of (5-2t+3t^2) is 63

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integral from 1 to 4 of (5-2t+3t^2)

\int_{1}^{4} (5 - 2 t + 3 t^{2}) d t

63

Solution steps

\int_{1}^{4} 5 - 2 t + 3 t^{2} 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_{1}^{4} 5 d t - \int_{1}^{4} 2 t d t + \int_{1}^{4} 3 t^{2} d t

\int_{1}^{4} 5 d t = 15

\int_{1}^{4} 2 t d t = 15

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

= 15 - 15 + 63

Simplify

= 63

\int 3 x^{2} cos (x^{3}) d x

f (x) = 5 e^{4 x}

x \frac{d y}{d x} + 2 y = - 3 e^{x}

\frac{d}{d x} (7 - 4 x)

\int 2 x^{- 4} d x

What is the integral from 1 to 4 of (5-2t+3t^2) ?
The integral from 1 to 4 of (5-2t+3t^2) is 63