Solve ∫ (from 0 to 10) of (0.3x^2-7.8x+48) wrt x

$$\int_{ 0 }^{ 10 } (0.3 { x }^{ 2 } -7.8x+48) d x$$

Evaluate

$$190$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{3x^{2}}{10}-\frac{39x}{5}+48\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{3x^{2}}{10}\mathrm{d}x+\int -\frac{39x}{5}\mathrm{d}x+\int 48\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{3\int x^{2}\mathrm{d}x}{10}-\frac{39\int x\mathrm{d}x}{5}+\int 48\mathrm{d}x$$

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

$$\frac{x^{3}}{10}-\frac{39\int x\mathrm{d}x}{5}+\int 48\mathrm{d}x$$

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}. Multiply -7.8 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{10}-\frac{39x^{2}}{10}+\int 48\mathrm{d}x$$

Find the integral of 48 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{10}-\frac{39x^{2}}{10}+48x$$

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.

$$\frac{10^{3}}{10}-\frac{39}{10}\times 10^{2}+48\times 10-\left(\frac{0^{3}}{10}-\frac{39}{10}\times 0^{2}+48\times 0\right)$$

Simplify.

$$190$$