Solve ∫ (8/7-1/2)^2+(3/4-3/7)^2

$$\int ( \frac { 8 } { 7 } - \frac { 1 } { 2 } ) ^ { 2 } + ( \frac { 3 } { 4 } - \frac { 3 } { 7 } ) ^ { 2 }$$

Evaluate

$$\frac{405x}{784}+С$$

Solution Steps

Subtract \frac{1}{2} from \frac{8}{7} to get \frac{9}{14}.

$$\int \left(\frac{9}{14}\right)^{2}+\left(\frac{3}{4}-\frac{3}{7}\right)^{2}\mathrm{d}x$$

Calculate \frac{9}{14} to the power of 2 and get \frac{81}{196}.

$$\int \frac{81}{196}+\left(\frac{3}{4}-\frac{3}{7}\right)^{2}\mathrm{d}x$$

Subtract \frac{3}{7} from \frac{3}{4} to get \frac{9}{28}.

$$\int \frac{81}{196}+\left(\frac{9}{28}\right)^{2}\mathrm{d}x$$

Calculate \frac{9}{28} to the power of 2 and get \frac{81}{784}.

$$\int \frac{81}{196}+\frac{81}{784}\mathrm{d}x$$

Add \frac{81}{196} and \frac{81}{784} to get \frac{405}{784}.

$$\int \frac{405}{784}\mathrm{d}x$$

Find the integral of \frac{405}{784} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{405x}{784}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{405x}{784}+С$$

Differentiate w.r.t. x

$$\frac{405}{784} = 0.5165816326530612$$