Solve ∫ (from - 1 to 7) of (1-(-x^2+2)^2)

$$\int _ { - 1 } ^ { 7 } ( 1 - ( - x ^ { 2 } + 2 ) ^ { 2 } )$$

Evaluate

$$-\frac{43904}{15}\approx -2926.933333333$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x^{2}+2\right)^{2}.

$$\int _{-1}^{7}1-\left(\left(-x^{2}\right)^{2}+4\left(-x^{2}\right)+4\right)\mathrm{d}x$$

Calculate -x^{2} to the power of 2 and get \left(x^{2}\right)^{2}.

$$\int _{-1}^{7}1-\left(\left(x^{2}\right)^{2}+4\left(-x^{2}\right)+4\right)\mathrm{d}x$$

To find the opposite of \left(x^{2}\right)^{2}+4\left(-x^{2}\right)+4, find the opposite of each term.

$$\int _{-1}^{7}1-\left(x^{2}\right)^{2}-4\left(-x^{2}\right)-4\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{-1}^{7}1-x^{4}-4\left(-x^{2}\right)-4\mathrm{d}x$$

Multiply -4 and -1 to get 4.

$$\int _{-1}^{7}1-x^{4}+4x^{2}-4\mathrm{d}x$$

Subtract 4 from 1 to get -3.

$$\int _{-1}^{7}-3-x^{4}+4x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -1 times \frac{x^{5}}{5}.

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

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

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

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.

$$-3\times 7-\frac{7^{5}}{5}+\frac{4}{3}\times 7^{3}-\left(-3\left(-1\right)-\frac{\left(-1\right)^{5}}{5}+\frac{4}{3}\left(-1\right)^{3}\right)$$

Simplify.

$$-\frac{43904}{15}$$