Q.Calculate the area of the region bounded by the curve $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the x-axis using integration.

Concept: Area under a curve using definite integration — the given equation is an ellipse in standard form.

The curve is $\frac{x^2}{9} + \frac{y^2}{4} = 1$, so solving for $y$ (upper half):

The ellipse meets the x-axis at $x = -3$ and $x = 3$. The required region is the upper half of the ellipse, so area is:

This is a standard integral: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$. Here $a = 3$:

$A = \frac{2}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3} \right]_{-3}^{3}$ …

The curve $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is an ellipse centered at the origin, with semi-major axis $a = 3$ along the $x$-axis and semi-minor axis $b = 2$ along the $y$-axis. The region bounded by this curve and the $x$-axis means we take only the upper half of the ellipse — the part where $y \ge 0$. The $x$-axis itself forms the lower boundary.

Why integration? The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is $\int_a^b f(x)\,dx$. Here, the upper boundary is the top half of the ellipse, and the lower boundary is $y = 0$. So we solve for $y$ from the ellipse equation, take the positive root, and integrate over the full horizontal span of the ellipse.

1. Solve for $y$ in terms of $x$. From $\frac{x^2}{9} + \frac{y^2}{4} = 1$, multiply through by $4$:

Taking the positive square root (upper half):

2. Determine the limits of integration.

   The ellipse meets the $x$-axis where $y = 0$, i.e., $\frac{x^2}{9} = 1$, so $x = \pm 3$. The region runs from $x = -3$ to $x = 3$.

3. Set up the area integral.

4. Evaluate the integral. The integral $\int \sqrt{a^2 - x^2}\,dx$ is a standard form. Here $a = 3$.

So:

Simplify the factor:

5. Plug in the limits. …