Solve ∫ {left(sqrt{x}-2sqrt{x}right)}^2 wrt x

$$\int{ { \left( \sqrt{ x } -2 \sqrt{ x } \right) }^{ 2 } }d x$$

Evaluate

$$\frac{x^{2}}{2}+С$$

Solution Steps

Combine \sqrt{x} and -2\sqrt{x} to get -\sqrt{x}.

$$\int \left(-\sqrt{x}\right)^{2}\mathrm{d}x$$

Expand \left(-\sqrt{x}\right)^{2}.

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

Calculate -1 to the power of 2 and get 1.

$$\int 1\left(\sqrt{x}\right)^{2}\mathrm{d}x$$

Calculate \sqrt{x} to the power of 2 and get x.

$$\int 1x\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}$$

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.

$$\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x$$