Solve ∫ (from 0 to 6) of 05x^2-3x wrt x

$$\int _ { 0 } ^ { 6 } 05 x ^ { 2 } - 3 x d x$$

Evaluate

$$-54$$

Solution Steps

Multiply 0 and 5 to get 0.

$$\int _{0}^{6}0x^{2}-3x\mathrm{d}x$$

Anything times zero gives zero.

$$\int _{0}^{6}0-3x\mathrm{d}x$$

Anything plus zero gives itself.

$$\int _{0}^{6}-3x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -3x\mathrm{d}x$$

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

$$-3\int x\mathrm{d}x$$

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}.

$$-\frac{3x^{2}}{2}$$

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.

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

Simplify.

$$-54$$