Solve ∫ (from 0 to 2) of (6.4x-0.64x^2) wrt x

$$\int_{ 0 }^{ 2 } (6.4x-0.64 { x }^{ 2 } ) d x$$

Evaluate

$$\frac{832}{75}\approx 11.093333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{32x}{5}-\frac{16x^{2}}{25}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{32x}{5}\mathrm{d}x+\int -\frac{16x^{2}}{25}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{32\int x\mathrm{d}x}{5}-\frac{16\int x^{2}\mathrm{d}x}{25}$$

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

$$\frac{16x^{2}}{5}-\frac{16\int x^{2}\mathrm{d}x}{25}$$

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 -0.64 times \frac{x^{3}}{3}.

$$\frac{16x^{2}}{5}-\frac{16x^{3}}{75}$$

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.

$$\frac{16}{5}\times 2^{2}-\frac{16}{75}\times 2^{3}-\left(\frac{16}{5}\times 0^{2}-\frac{16}{75}\times 0^{3}\right)$$

Simplify.

$$\frac{832}{75}$$