Solve ∫ (from - 1 to 2) of y^2y-2 wrt y

$$\int _ { - 1 } ^ { 2 } y ^ { 2 } y - 2 d y$$

Evaluate

$$-\frac{9}{4}=-2.25$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int _{-1}^{2}y^{3}-2\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int y^{3}-2\mathrm{d}y$$

Integrate the sum term by term.

$$\int y^{3}\mathrm{d}y+\int -2\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{3}\mathrm{d}y with \frac{y^{4}}{4}.

$$\frac{y^{4}}{4}+\int -2\mathrm{d}y$$

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{y^{4}}{4}-2y$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{2^{4}}{4}-2\times 2-\left(\frac{\left(-1\right)^{4}}{4}-2\left(-1\right)\right)$$

Simplify.

$$-\frac{9}{4}$$