Solve ∫ (from 0 to 4) of (-3.6x)(-0.1x)/2 wrt x

$$\int_{ 0 }^{ 4 } \frac{ (-3.6x)(-0.1x) }{ 2 } d x$$

Evaluate

$$3.84$$

Solution Steps

Multiply x and x to get x^{2}.

$$\int _{0}^{4}\frac{-3.6x^{2}\left(-0.1\right)}{2}\mathrm{d}x$$

Multiply -3.6 and -0.1 to get 0.36.

$$\int _{0}^{4}\frac{0.36x^{2}}{2}\mathrm{d}x$$

Divide 0.36x^{2} by 2 to get 0.18x^{2}.

$$\int _{0}^{4}0.18x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{9x^{2}}{50}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\frac{9\int x^{2}\mathrm{d}x}{50}$$

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}.

$$\frac{3x^{3}}{50}$$

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.

$$\frac{3}{50}\times 4^{3}-\frac{3}{50}\times 0^{3}$$

Simplify.

$$\frac{96}{25}$$