Solve ∫ (from 0 to 3) of 9-x^2-x^2-1 wrt x

$$\int _ { 0 } ^ { 3 } 9 - x ^ { 2 } - x ^ { 2 } - 1 d x$$

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 9-x^{2}-x^{2}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}x-\int x^{2}\mathrm{d}x-\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-\int x^{2}\mathrm{d}x-\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

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 -1 times \frac{x^{3}}{3}.

$$9x-\frac{x^{3}}{3}-\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

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 -1 times \frac{x^{3}}{3}.

$$9x-\frac{x^{3}}{3}-\frac{x^{3}}{3}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-\frac{x^{3}}{3}-\frac{x^{3}}{3}-x$$

Simplify.

$$8x-\frac{2x^{3}}{3}$$

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.

$$8\times 3-\frac{2}{3}\times 3^{3}-\left(8\times 0-\frac{2}{3}\times 0^{3}\right)$$

Simplify.

$$6$$