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

The integral from 0 to 4 of (4t-1)^2 is 844/3

Popular >

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

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

\frac{844}{3}

Decimal

281.33333 \dots

Solution steps

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

Expand

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

= \int_{0}^{4} 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}^{4} 16 t^{2} d t - \int_{0}^{4} 8 t d t + \int_{0}^{4} 1 d t

\int_{0}^{4} 16 t^{2} d t = \frac{1024}{3}

\int_{0}^{4} 8 t d t = 64

\int_{0}^{4} 1 d t = 4

= \frac{1024}{3} - 64 + 4

Simplify

\frac{1024}{3} - 64 + 4 : \frac{844}{3}

= \frac{844}{3}

\int_{- 1}^{4} (x + 4) - (x^{2} - 2 x) d x

\int_{\frac{π}{12}}^{\frac{π}{9}} 5 x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

\int_{0}^{1} x^{\frac{5}{6}} d x

\int_{0}^{1} (x + 2 1 - x^{2}) d x

\int_{0}^{1} \frac{1}{4 + 5 x - 1} d x

What is the integral from 0 to 4 of (4t-1)^2 ?
The integral from 0 to 4 of (4t-1)^2 is 844/3