Solve ∫ (from 0 to sqrt { 2) of }(6t^2-3t^4)dt

$$\int _ { 0 } ^ { \sqrt { 2 } } ( 6 t ^ { 2 } - 3 t ^ { 4 } ) d t$$

Evaluate

$$\frac{8\sqrt{2}}{5}\approx 2.2627417$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6t^{2}-3t^{4}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 6t^{2}\mathrm{d}t+\int -3t^{4}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$6\int t^{2}\mathrm{d}t-3\int t^{4}\mathrm{d}t$$

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 6 times \frac{t^{3}}{3}.

$$2t^{3}-3\int t^{4}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{4}\mathrm{d}t with \frac{t^{5}}{5}. Multiply -3 times \frac{t^{5}}{5}.

$$2t^{3}-\frac{3t^{5}}{5}$$

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.

$$2\times \left(2^{\frac{1}{2}}\right)^{3}-\frac{3}{5}\times \left(2^{\frac{1}{2}}\right)^{5}-\left(2\times 0^{3}-\frac{3}{5}\times 0^{5}\right)$$

Simplify.

$$\frac{8\sqrt{2}}{5}$$