Solve ∫ (from - 1 to 4) of (x^2+2x-5) wrt x

$$\int _ { - 1 } ^ { 4 } ( x ^ { 2 } + 2 x - 5 ) d x$$

Evaluate

$$\frac{35}{3}\approx 11.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

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

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

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

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

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

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.

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

Simplify.

$$\frac{35}{3}$$