Solve ∫ (from 0 to 1) of frac{2^{-1}-2^2-2^-1}{2^2+2^2}

$$\int _ { 0 } ^ { 1 } \frac { 2 ^ { - 1 } - 2 ^ { 2 } - 2 ^ { - 1 } } { 2 ^ { 2 } + 2 ^ { 2 } }$$

Evaluate

$$-\frac{1}{2}=-0.5$$

Solution Steps

Calculate 2 to the power of -1 and get \frac{1}{2}.

$$\int _{0}^{1}\frac{\frac{1}{2}-2^{2}-2^{-1}}{2^{2}+2^{2}}\mathrm{d}x$$

Calculate 2 to the power of 2 and get 4.

$$\int _{0}^{1}\frac{\frac{1}{2}-4-2^{-1}}{2^{2}+2^{2}}\mathrm{d}x$$

Subtract 4 from \frac{1}{2} to get -\frac{7}{2}.

$$\int _{0}^{1}\frac{-\frac{7}{2}-2^{-1}}{2^{2}+2^{2}}\mathrm{d}x$$

Calculate 2 to the power of -1 and get \frac{1}{2}.

$$\int _{0}^{1}\frac{-\frac{7}{2}-\frac{1}{2}}{2^{2}+2^{2}}\mathrm{d}x$$

Subtract \frac{1}{2} from -\frac{7}{2} to get -4.

$$\int _{0}^{1}\frac{-4}{2^{2}+2^{2}}\mathrm{d}x$$

Calculate 2 to the power of 2 and get 4.

$$\int _{0}^{1}\frac{-4}{4+2^{2}}\mathrm{d}x$$

Calculate 2 to the power of 2 and get 4.

$$\int _{0}^{1}\frac{-4}{4+4}\mathrm{d}x$$

Add 4 and 4 to get 8.

$$\int _{0}^{1}\frac{-4}{8}\mathrm{d}x$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$\int _{0}^{1}-\frac{1}{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{1}{2}\mathrm{d}x$$

Find the integral of -\frac{1}{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x}{2}$$

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.

$$-\frac{1}{2}+\frac{1}{2}\times 0$$

Simplify.

$$-\frac{1}{2}$$