Solve ∫ (from 0 to 4) of 1,5x-4 wrt x

$$\int _ { 0 } ^ { 4 } 1,5 x - 4 d x$$

Evaluate

$$-4$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 1,5 times \frac{x^{2}}{2}.

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{3}{4}\times 4^{2}-4\times 4-\left(\frac{3}{4}\times 0^{2}-4\times 0\right)$$

Simplify.

$$-4$$