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Popular >

integral of (2t-1)^2

Solution

\int (2 t - 1)^{2} d t

Solution

\frac{4 t ^{3}}{3} - 2 t^{2} + t + C

Solution steps

\int (2 t - 1)^{2} d t

Expand

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

= \int 4 t^{2} - 4 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 4 t^{2} d t - \int 4 t d t + \int 1 d t

\int 4 t^{2} d t = \frac{4 t ^{3}}{3}

\int 4 t d t = 2 t^{2}

\int 1 d t = t

= \frac{4 t ^{3}}{3} - 2 t^{2} + t

Add a constant to the solution

= \frac{4 t ^{3}}{3} - 2 t^{2} + t + C

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Frequently Asked Questions (FAQ)

What is the integral of (2t-1)^2 ?
The integral of (2t-1)^2 is (4t^3)/3-2t^2+t+C