Solve = ∫ (from 0 to 4) of ((-1*5*2.5)+40x))*xx) wrt x

$$= \int_{ 0 }^{ 4 } ((-1 \times 5 \times 2.5)+40x)) \times xx) d x$$

Evaluate

$$270$$

Solution Steps

Multiply -1 and 5 to get -5.

$$\int _{0}^{4}-5\times 2.5+40x\mathrm{d}x$$

Multiply -5 and 2.5 to get -12.5.

$$\int _{0}^{4}-12.5+40x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -12.5+40x\mathrm{d}x$$

Integrate the sum term by term.

$$\int -12.5\mathrm{d}x+\int 40x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -12.5\mathrm{d}x+40\int x\mathrm{d}x$$

Find the integral of -12.5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{25x}{2}+40\int x\mathrm{d}x$$

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 40 times \frac{x^{2}}{2}.

$$-\frac{25x}{2}+20x^{2}$$

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.

$$-12.5\times 4+20\times 4^{2}-\left(-12.5\times 0+20\times 0^{2}\right)$$

Simplify.

$$270$$