Solve ∫ (from - 1 to 2) of -y^4-y^3+3/2y^2+2y

$$\int _ { - 1 } ^ { 2 } - y ^ { 4 } - y ^ { 3 } + \frac { 3 } { 2 } y ^ { 2 } + 2 y$$

Evaluate

$$\frac{3y\left(4+3y-2y^{2}-2y^{3}\right)}{2}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -y^{4}-y^{3}+\frac{3y^{2}}{2}+2y\mathrm{d}x$$

Find the integral of -y^{4}-y^{3}+\frac{3y^{2}}{2}+2y using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\left(-y^{4}-y^{3}+\frac{3y^{2}}{2}+2y\right)x$$

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.

$$\left(-y^{4}-y^{3}+\frac{3}{2}y^{2}+2y\right)\times 2-\left(-y^{4}-y^{3}+\frac{3}{2}y^{2}+2y\right)\left(-1\right)$$

Simplify.

$$\frac{3y\left(-2y^{3}-2y^{2}+3y+4\right)}{2}$$

Differentiate w.r.t. y

$$3\left(y+1\right)\left(2+y-4y^{2}\right)$$