What is the integral from 0 to 5 of (4t-1)^2 ?

The integral from 0 to 5 of (4t-1)^2 is 1715/3

integral from 0 to 5 of (4t-1)^2

\int_{0}^{5} (4 t - 1)^{2} d t

\frac{1715}{3}

Decimal

571.66666 \dots

Solution steps

\int_{0}^{5} (4 t - 1)^{2} d t

Expand

(4 t - 1)^{2} : 16 t^{2} - 8 t + 1

= \int_{0}^{5} 16 t^{2} - 8 t + 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_{0}^{5} 16 t^{2} d t - \int_{0}^{5} 8 t d t + \int_{0}^{5} 1 d t

\int_{0}^{5} 16 t^{2} d t = \frac{2000}{3}

\int_{0}^{5} 8 t d t = 100

\int_{0}^{5} 1 d t = 5

= \frac{2000}{3} - 100 + 5

Simplify

\frac{2000}{3} - 100 + 5 : \frac{1715}{3}

= \frac{1715}{3}

\int_{0}^{1} \frac{5 x ^{2}}{2 x ^{3} + 1} d x

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

\int_{0}^{x} sin (2 t^{2}) d t

What is the integral from 0 to 5 of (4t-1)^2 ?
The integral from 0 to 5 of (4t-1)^2 is 1715/3