Solve ∫ (from -1 to 0) of (x^3-3x+5) wrt x

$$\int_{ -1 }^{ 0 } ( { x }^{ 3 } -3x+5) d x$$

Evaluate

$$\frac{25}{4}=6.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-3x+5\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -3x\mathrm{d}x+\int 5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-3\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-3\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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 -3 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{4}-\frac{3x^{2}}{2}+\int 5\mathrm{d}x$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-\frac{3x^{2}}{2}+5x$$

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.

$$\frac{0^{4}}{4}-\frac{3}{2}\times 0^{2}+5\times 0-\left(\frac{\left(-1\right)^{4}}{4}-\frac{3}{2}\left(-1\right)^{2}+5\left(-1\right)\right)$$

Simplify.

$$\frac{25}{4}$$