Solve ∫ (from 0 to 4) of (-25+36x)(x) wrt x

$$\int_{ 0 }^{ 4 } (-25+36x)(x) d x$$

Evaluate

$$568$$

Solution Steps

Use the distributive property to multiply -25+36x by x.

$$\int _{0}^{4}-25x+36x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -25x+36x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -25x\mathrm{d}x+\int 36x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-25\int x\mathrm{d}x+36\int x^{2}\mathrm{d}x$$

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 -25 times \frac{x^{2}}{2}.

$$-\frac{25x^{2}}{2}+36\int x^{2}\mathrm{d}x$$

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 36 times \frac{x^{3}}{3}.

$$-\frac{25x^{2}}{2}+12x^{3}$$

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.

$$-\frac{25}{2}\times 4^{2}+12\times 4^{3}-\left(-\frac{25}{2}\times 0^{2}+12\times 0^{3}\right)$$

Simplify.

$$568$$