Q.Find the area of the smaller region bounded by the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the line $\dfrac{x}{a}+\dfrac{y}{b}=1$. OR Prove that $\displaystyle\int_0^{\pi/2}\log(\sin x)\,dx=-\dfrac{\pi}{2}\log 2$.

The core concept here is using definite integrals to calculate the area under a curve. When we have a region bounded by a curve $y=f(x)$, the x-axis, and vertical lines $x=a$ and $x=b$, the area is given by $\int_a^b f(x)\, dx$. Similarly, if the region is bounded by a curve $x=g(y)$, the y-axis, and horizontal lines $y=c$ and $y=d$, the area is $\int_c^d g(y)\, dy$.

The expressions in the options, $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$, immediately point to the equation of a circle. The general equation of a circle centered at the origin with radius $r$ is $x^2+y^2=r^2$. Comparing this with $9-x^2$ or $9-y^2$, we see that $r^2=9$, which means the radius $r=3$.

For the upper half of this circle, we can express $y$ as a function of $x$: $y^2 = 9-x^2 \implies y = \sqrt{9-x^2}$ (taking the positive root for the upper half).

For the right half of this circle, we can express $x$ as a function of $y$: $x^2 = 9-y^2 \implies x = \sqrt{9-y^2}$ (taking the positive root for the right half).

Since the figure is not provided, we must infer the shaded region from the given options. Options (C) and (D) involve integration from $0$ to $3$, which is the radius of the circle. This strongly suggests that the shaded region is either a quarter circle or a semi-circle.

1. Identify the circle's equation and radius:

   The terms $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ indicate that the circle has the equation $x^2+y^2=9$. This is a circle centered at the origin $(0,0)$ with a radius $r=3$.

2. Interpret the shaded region based on options:

   • Option (C) is $\int_{0}^{3} \sqrt{9-x^2}\, dx$. This integral represents the area under the curve $y=\sqrt{9-x^2}$ (the upper semi-circle) from $x=0$ to $x=3$. This region is precisely the quarter circle located in the first quadrant.
   • Option (D) is $2 \int_{0}^{3} \sqrt{9-x^2}\, dx$. This would be twice the area of the quarter circle, meaning it represents the area of the entire upper semi-circle (from $x=-3$ to $x=3$, or by symmetry, $2 \times$ area from $x=0$ to $x=3$).
   • Option (A) is $\int_{1}^{3} \sqrt{9-y^2}\, dy$. This represents the area under the curve $x=\sqrt{9-y^2}$ (the right semi-circle) from $y=1$ to $y=3$. This is a specific segment of the quarter circle, not the entire quarter circle.
   • Option (B) is $2 \int_{1}^{3} \sqrt{9-y^2}\, dy$. This would be twice the area in (A), representing a horizontal strip of the circle symmetric about the y-axis. …

The problem asks for the area enclosed by the curve $y = \sqrt{x}$, the vertical lines $x = 0$ and $x = 4$, and the x-axis. This is a classic "area under a curve" problem — the region is bounded above by the curve, below by the x-axis, and on the sides by two vertical lines. The key idea is that the area between a curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$, provided $f(x) \ge 0$ on that interval. Here, $\sqrt{x}$ is non-negative for $x \ge 0$, so we can directly integrate.

Let’s work through the calculation step by step.

1. Set up the integral. The region is bounded by $x = 0$ on the left and $x = 4$ on the right. The curve is $y = \sqrt{x}$, and the lower boundary is the x-axis ($y = 0$). So the area $A$ is:

2. Rewrite the integrand. Recall that $\sqrt{x} = x^{1/2}$. This makes the power rule for integration straightforward:

3. Apply the power rule. …