Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

The area of the region $\left\{(x,y)\in {\mathrm{ℝ}}^2:y^2\le x, x=4\right\}$ is- 

If the area of the triangle formed by the straight line $\frac{x}{2}+\frac{y}{3}=1$ with the coordinate axes is $k$ sq. units, then the value of $k$ is

The area of the region in the first quadrant, enclosed by the parabola $y^2=16x$ and its latus-rectum is $\frac{k}{3}$.

Find the value of $k$.

Find the area of the region bounded by $y=x^2$ and $y=4$. (Write the answer in fraction and in $\mathrm{sq}. \mathrm{units}$)

Find the area bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Find by integration the area of the region $\left\{\left(x, y\right)\in {\mathrm{ℝ}}^2:\frac{x^2}{9}+\frac{y^2}{4}\le 1\le \frac{x}{3}+\frac{y}{2}\right\}$.

Find the area of the smaller portion of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a^2>b^2$ cut off by the line $\frac{x}{a}+\frac{y}{b}=1$.

Determine the area of the region given by $\left\{\left(x, y\right)\in {\mathrm{ℝ}}^2:\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1\le \frac{x}{a}+\frac{y}{b}, a^2>b^2\right\}$.