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

The integral from-2 to 1 of (1/3 t-2)^2 is 43/3

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

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

\frac{43}{3}

Solution steps

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

Apply u-substitution

: \int_{- \frac{8}{3}}^{- \frac{5}{3}} u^{2} \cdot 3 d u

= \int_{- \frac{8}{3}}^{- \frac{5}{3}} u^{2} \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int_{- \frac{8}{3}}^{- \frac{5}{3}} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- \frac{8}{3}}^{- \frac{5}{3}}

= 3 [\frac{u ^{3}}{3}]_{- \frac{8}{3}}^{- \frac{5}{3}}

Compute the boundaries:

\frac{43}{9}

= 3 \cdot \frac{43}{9}

Simplify

= \frac{43}{3}

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What is the integral from-2 to 1 of (1/3 t-2)^2 ?
The integral from-2 to 1 of (1/3 t-2)^2 is 43/3