Solve ∫ (from 0 to 2) of (1.5x)(-1/3x) wrt x

$$\int_{ 0 }^{ 2 } (1.5x)(- \frac{ 1 }{ 3 } x) d x$$

Evaluate

$$-\frac{4}{3}\approx -1.333333333$$

Solution Steps

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

$$\int _{0}^{2}1.5x^{2}\left(-\frac{1}{3}\right)\mathrm{d}x$$

Convert decimal number 1.5 to fraction \frac{15}{10}. Reduce the fraction \frac{15}{10} to lowest terms by extracting and canceling out 5.

$$\int _{0}^{2}\frac{3}{2}x^{2}\left(-\frac{1}{3}\right)\mathrm{d}x$$

Multiply \frac{3}{2} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$\int _{0}^{2}\frac{3\left(-1\right)}{2\times 3}x^{2}\mathrm{d}x$$

Cancel out 3 in both numerator and denominator.

$$\int _{0}^{2}\frac{-1}{2}x^{2}\mathrm{d}x$$

Fraction \frac{-1}{2} can be rewritten as -\frac{1}{2} by extracting the negative sign.

$$\int _{0}^{2}-\frac{1}{2}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

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

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

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}.

$$-\frac{x^{3}}{6}$$

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.

$$-\frac{2^{3}}{6}+\frac{0^{3}}{6}$$

Simplify.

$$-\frac{4}{3}$$