Solve ∫ (from 0 to 1) of 2^2(2x)*(2x) wrt x

$$\int _ { 0 } ^ { 1 } 2 ^ { 2 } ( 2 x ) \cdot ( 2 x ) d x$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int _{0}^{1}2^{3}x\times 2x\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\int _{0}^{1}2^{4}xx\mathrm{d}x$$

Multiply x and x to get x^{2}.

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

Calculate 2 to the power of 4 and get 16.

$$\int _{0}^{1}16x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 16x^{2}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$16\int x^{2}\mathrm{d}x$$

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}.

$$\frac{16x^{3}}{3}$$

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.

$$\frac{16}{3}\times 1^{3}-\frac{16}{3}\times 0^{3}$$

Simplify.

$$\frac{16}{3}$$