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Popular >

integral from 0 to 3 of (4t-1)^2

Solution

\int_{0}^{3} (4 t - 1)^{2} d t

Solution

111

Solution steps

\int_{0}^{3} (4 t - 1)^{2} d t

Expand

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

= \int_{0}^{3} 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}^{3} 16 t^{2} d t - \int_{0}^{3} 8 t d t + \int_{0}^{3} 1 d t

\int_{0}^{3} 16 t^{2} d t = 144

\int_{0}^{3} 8 t d t = 36

\int_{0}^{3} 1 d t = 3

= 144 - 36 + 3

Simplify

= 111

Graph

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\int_{0}^{\infty} e^{- 4 x^{2}} d x

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

Frequently Asked Questions (FAQ)

What is the integral from 0 to 3 of (4t-1)^2 ?
The integral from 0 to 3 of (4t-1)^2 is 111