Solve ∫ (from 0.50 to 1.00) of (52.8x+38.4x^2) wrt x

$$\int _ { 0.50 } ^ { 1.00 } ( 52.8 x + 38.4 x ^ { 2 } ) d x$$

Evaluate

$$31$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{264x+192x^{2}}{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{264x}{5}\mathrm{d}x+\int \frac{192x^{2}}{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{264\int x\mathrm{d}x+192\int x^{2}\mathrm{d}x}{5}$$

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

$$\frac{132x^{2}+192\int x^{2}\mathrm{d}x}{5}$$

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

$$\frac{132x^{2}+64x^{3}}{5}$$

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.

$$\frac{132}{5}\times 1^{2}+\frac{64}{5}\times 1^{3}-\left(\frac{132}{5}\times 0.5^{2}+\frac{64}{5}\times 0.5^{3}\right)$$

Simplify.

$$31$$