Solve ∫ (from z to 3) of 4x^2+3x wrt x

$$\int _ { z } ^ { 3 } 4 x ^ { 2 } + 3 x d x$$

Evaluate

$$-\frac{4z^{3}}{3}-\frac{3z^{2}}{2}+\frac{99}{2}$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 4 times \frac{x^{3}}{3}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 3 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}+\frac{3x^{2}}{2}$$

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.

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

Simplify.

$$-\frac{3z^{2}}{2}-\frac{4z^{3}}{3}+\frac{99}{2}$$

Differentiate w.r.t. z

$$z\left(-4z-3\right)$$