# Find the area under the curve by using integration:

Find the area under the curve by using integration: My Attempt: Considering a strip of width $$dx$$ and height $$y$$ at a distance of $$x$$ from $$Y-$$ axis. $$A=\int_{0}^{a} y dx$$ $$=\int_{0}^{a} k(x-a)^{2} dx$$ $$=\int_{0}^{a} \frac {b}{a^2} (x^2-2ax+a^2)dx$$ $$=\frac {b}{a^2} \int_{0}^{a} (x^2-2ax+a^2) dx$$ Thus Area$$=\frac {ab}{3}$$

Now if we consider a strip of width $$dy$$ parallel to $$X-$$ axis $$A=\int_{0}^{b} xdy$$ $$=\int_{0}^{b} \sqrt {\frac {y}{k}}+a dy$$ $$=\frac {1}{\sqrt {k}} (\frac {2}{3} b^{\frac {3}{2}})+ab$$ Thus Area$$=\frac {5ab}{3}$$

Why am I getting different answers?

• Because the inverse function of $x\mapsto k(x-a)^2$ is $a-\sqrt{\frac{y}{b}}$ and not what you wrote. Remember that $\sqrt{x^2}\neq |x|$ (and not $x$). – Surb Mar 31 '20 at 9:24

The reason you're getting different answers is you took the wrong sign when you converted the equation of

$$y = k(x-a)^2 \tag{1}\label{eq1A}$$

Note that in the region you're integrating, $$x \le a \implies x - a \le 0$$. Thus, when you took the square of both sides in your change of variables, you should have used the negative value instead to get

$$x = -\sqrt{\frac{y}{k}} + a \tag{2}\label{eq2A}$$

If you use that, then your integration will work properly. In particular, you would then get

\begin{equation}\begin{aligned} A & = \int_{0}^{b} \left(-\sqrt{\frac{y}{k}} + a\right)dy \\ & = -\frac{1}{\sqrt {k}}\left(\frac {2}{3} b^{\frac {3}{2}}\right)+ab \\ & = -\frac{2ab}{3} + ab \\ & = \frac{ab}{3} \end{aligned}\end{equation}\tag{3}\label{eq3A}

• However the correct answer is $\frac {ab}{3}$. Yes? – pi-π Mar 31 '20 at 9:56
• @pi-π Yes, the correct answer is $\frac{ab}{3}$. You got that using your first method, and with the correction given in my answer to your second method, you would then also get the same result of $\frac{ab}{3}$. – John Omielan Mar 31 '20 at 9:57
• Thank You....... – pi-π Mar 31 '20 at 9:59