Question

Integrate with respect to t :
∫04t√(t−2)dt

Answer 1

Answer on Question #50939 - Math - Integral Calculus

Integrate with respect to $t$:

$$\int_{0}^{4} t \sqrt{t - 2} \, dt$$

Solution

The domain of the integrand is given by

$$t \in [2, +\infty],$$

since the square root of negative number doesn't exist (at least in real analysis). Therefore the given integral doesn't exist, since the integrand is integrated over the region $[0, 2)$ where it's not defined.

**Answer**: doesn't exist.

2) Using Newton-Leibnitz formula and integral formula for powers,

$$\begin{aligned} \int_{2}^{4} t \sqrt{t - 2} \, dt &= \int_{2}^{4} (t - 2) \sqrt{t - 2} \, dt + 2 \int_{2}^{4} \sqrt{t - 2} \, dt = |u := t - 2, \, du = dt| \\ &= \int_{0}^{2} u \sqrt{u} \, du + 2 \int_{0}^{2} \sqrt{u} \, du = \left. \frac{u^{5/2}}{5/2} \right|_{0}^{2} + 2 \left. \frac{u^{3/2}}{3/2} \right|_{0}^{2} = \frac{2}{5} \cdot 4\sqrt{2} + \frac{4}{3} \cdot 2\sqrt{2} = \frac{64\sqrt{2}}{15} \end{aligned}$$

3)

$$\int 0.4t \sqrt{t - 2} \, dt = 0.4 \int t \sqrt{t - 2} \, dt = 0.4 \int (t - 2) \sqrt{t - 2} \, dt + 0.4 \cdot 2 \int \sqrt{t - 2} \, dt = \\ |u := t - 2, \, du = dt| = 0.4 \int u \sqrt{u} \, du + 0.8 \int \sqrt{u} \, du = 0.4 \frac{u^{5/2}}{5/2} + 0.8 \frac{u^{3/2}}{3/2} + C, \text{ where } C \text{ is an arbitrary real constant}.$$

4)

$$\int 0.4(t - 2)\sqrt{t} \, dt = 0.4 \int t \sqrt{t} \, dt - 0.4 \cdot 2 \int \sqrt{t} \, dt = 0.4 \int t \sqrt{t} \, dt - 0.8 \int \sqrt{t} \, dt = 0.4 \frac{t^{5/2}}{5/2} - 0.8 \frac{u^{\frac{3}{2}}}{2} + C, \text{ where } C \text{ is an arbitrary real constant}.$$

5)

$$\int_{0}^{4} (t - 2)\sqrt{t} \, dt = \int_{0}^{4} t \sqrt{t} \, dt - 2 \int_{0}^{4} \sqrt{t} \, dt = \left. \frac{t^{5/2}}{5/2} \right|_{0}^{4} - 2 \left. \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right|_{0}^{4} = \frac{2}{5} \cdot 32 - \frac{4}{3} \cdot 8 = \frac{32}{15}$$

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