How to integrate $ \int_{t=0}^t \sqrt{2+2.25t} \, dt = 1 $?

Question

While solving a physics problem I was confronted by this equation $$ \int_{t=0}^t \sqrt{2+2.25t}\,\mathrm dt = 1 $$ The answer is $t = 0.612$ seconds. How do I integrate this? Would someone please explain to me in simple terms how I can solve this without spending a lot of time (more than two-three minutes) over this?

Answer 1

I'll change the dummy variable to avoid confusion. The integral is

$$ \int_{\tau=0}^{\tau=t} \sqrt{2 + \frac94\tau}\ d\tau $$

Substitute $u = 2 + \frac94\tau$ we get

$$ \frac49\int_{u=2}^{u=2+9t/4} \sqrt{u}\ du = \frac{8}{27} u^{3/2}\Bigg|_{u=2}^{u=2+9t/4} = \frac{8}{27} \left[\left(2+\frac94 t\right)^{3/2}- 2\sqrt{2}\right] $$

You can solve this for $t$

\begin{align} \left(2+\frac94 t\right)^{3/2} - 2\sqrt{2} &= \frac{27}{8} \\ \left(2+\frac94 t\right)^{3/2} &= \frac{27}{8}+ 2\sqrt{2} \\ 2 + \frac94 t &= \left(\frac{27}{8} + 2\sqrt{2}\right)^{2/3} \\ &\cdots \end{align}

Answer 2

$$\int_0^t \sqrt{a+b x} \ \mathrm{d}x$$

Substitute $u=a+bx$ so $\dfrac{du}{dx}=b$:

$$\int_a^{a+t b} \frac{1}{b} \sqrt{u} \mathrm{d}u$$

I think you can do $\int \sqrt{t} \ \mathrm{d}t$.