Solve ∫ (from 0 to 2) of (24+24x+06x^2)(x) wrt x

$$\int_{ 0 }^{ 2 } (24+24x+06 { x }^{ 2 } )(x) d x$$

Evaluate

$$112$$

Solution Steps

Multiply 0 and 6 to get 0.

$$\int _{0}^{2}\left(24+24x+0x^{2}\right)x\mathrm{d}x$$

Anything times zero gives zero.

$$\int _{0}^{2}\left(24+24x+0\right)x\mathrm{d}x$$

Add 24 and 0 to get 24.

$$\int _{0}^{2}\left(24+24x\right)x\mathrm{d}x$$

Use the distributive property to multiply 24+24x by x.

$$\int _{0}^{2}24x+24x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 24x+24x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 24x\mathrm{d}x+\int 24x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$24\int x\mathrm{d}x+24\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 24 times \frac{x^{2}}{2}.

$$12x^{2}+24\int x^{2}\mathrm{d}x$$

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

$$12x^{2}+8x^{3}$$

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.

$$12\times 2^{2}+8\times 2^{3}-\left(12\times 0^{2}+8\times 0^{3}\right)$$

Simplify.

$$112$$