Solve ∫ (from 0 to x) of (18t^2+sqrt{t})dt

$$\int _ { 0 } ^ { x } ( 18 t ^ { 2 } + \sqrt { t } ) d t$$

Evaluate

$$6x^{3}+\frac{2x^{\frac{3}{2}}}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 18t^{2}+\sqrt{t}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 18t^{2}\mathrm{d}t+\int \sqrt{t}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$18\int t^{2}\mathrm{d}t+\int \sqrt{t}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply 18 times \frac{t^{3}}{3}.

$$6t^{3}+\int \sqrt{t}\mathrm{d}t$$

Rewrite \sqrt{t} as t^{\frac{1}{2}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{\frac{1}{2}}\mathrm{d}t with \frac{t^{\frac{3}{2}}}{\frac{3}{2}}. Simplify.

$$6t^{3}+\frac{2t^{\frac{3}{2}}}{3}$$

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.

$$6x^{3}+\frac{2}{3}x^{\frac{3}{2}}-\left(6\times 0^{3}+\frac{2}{3}\times 0^{\frac{3}{2}}\right)$$

Simplify.

$$6x^{3}+\frac{2x^{\frac{3}{2}}}{3}$$

Differentiate w.r.t. x

$$18x^{2}+\sqrt{x}$$