What is the integral from 1 to 2 of (64/3-8y^2) ?

The integral from 1 to 2 of (64/3-8y^2) is 8/3

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integral from 1 to 2 of (64/3-8y^2)

\int_{1}^{2} (\frac{64}{3} - 8 y^{2}) d y

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

\int_{1}^{2} \frac{64}{3} - 8 y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{64}{3} d y - \int_{1}^{2} 8 y^{2} d y

\int_{1}^{2} \frac{64}{3} d y = \frac{64}{3}

\int_{1}^{2} 8 y^{2} d y = \frac{56}{3}

= \frac{64}{3} - \frac{56}{3}

Simplify

\frac{64}{3} - \frac{56}{3} : \frac{8}{3}

= \frac{8}{3}

\int_{- 2}^{3} (5 x^{2} - 3 x - 2)^{2} d x

\int_{0}^{6} ∣ x - 2 ∣ d x

\int_{0}^{1} - x^{2} - 3 x d x

\int_{1}^{3} (x^{3} + 1) d x

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

What is the integral from 1 to 2 of (64/3-8y^2) ?
The integral from 1 to 2 of (64/3-8y^2) is 8/3