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

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

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

\int_{1}^{2} 4 π (t - 2)^{2} d t

\frac{4 π}{3}

Decimal

4.18879 \dots

Solution steps

\int_{1}^{2} 4 π (t - 2)^{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 π \cdot \int_{1}^{2} (t - 2)^{2} d t

Expand

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

= 4 π \cdot \int_{1}^{2} t^{2} - 4 t + 4 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

= 4 π (\int_{1}^{2} t^{2} d t - \int_{1}^{2} 4 t d t + \int_{1}^{2} 4 d t)

\int_{1}^{2} t^{2} d t = \frac{7}{3}

\int_{1}^{2} 4 t d t = 6

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

= 4 π (\frac{7}{3} - 6 + 4)

Simplify

4 π (\frac{7}{3} - 6 + 4) : \frac{4 π}{3}

= \frac{4 π}{3}

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