Solve ∫ (from 0 to 4) of (-7.4x+x^2)(-0.1x) wrt x

$$\int_{ 0 }^{ 4 } (-7.4x+ { x }^{ 2 } )(-0.1x) d x$$

Evaluate

$$\frac{704}{75}\approx 9.386666667$$

Solution Steps

Use the distributive property to multiply -7.4x+x^{2} by -0.1.

$$\int _{0}^{4}\left(0.74x-0.1x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply 0.74x-0.1x^{2} by x.

$$\int _{0}^{4}0.74x^{2}-0.1x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{37x^{2}}{50}-\frac{x^{3}}{10}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{37x^{2}}{50}\mathrm{d}x+\int -\frac{x^{3}}{10}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{37\int x^{2}\mathrm{d}x}{50}-\frac{\int x^{3}\mathrm{d}x}{10}$$

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

$$\frac{37x^{3}}{150}-\frac{\int x^{3}\mathrm{d}x}{10}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -0.1 times \frac{x^{4}}{4}.

$$\frac{37x^{3}}{150}-\frac{x^{4}}{40}$$

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.

$$\frac{37}{150}\times 4^{3}-\frac{4^{4}}{40}-\left(\frac{37}{150}\times 0^{3}-\frac{0^{4}}{40}\right)$$

Simplify.

$$\frac{704}{75}$$