Solve the Following Question.(3 Marks)

Question 13 Marks

Find the area of the region bounded by the parabola $y^2=25 x$ and the line $x=5$.

Answer

Given the equation of the parabola is $y^2=25 x$ $\therefore y =5 \sqrt{ } x \ldots \ldots[\because \because$ In first quadrant, $y >0]$
Required area $=$ area of the region OQRPO $=2($ area of the region ORPO)
$
\begin{aligned}
& =2 \int_0^5 y \cdot d x \\
& =2 \int_0^5 5 \sqrt{x} \cdot d x \\
& =10 \int_0^5 x^{\frac{1}{2}} \cdot d x \\
& =10\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^5 \\
& =\frac{20}{3}\left[(5)^{\frac{3}{2}}-0\right] \\
& =\frac{20}{3}(5 \sqrt{(5)} \\
& =\frac{100 \sqrt{5}}{3} \text { sq.units. }
\end{aligned}
$

View full question & answer→

Question 23 Marks

Find the area of the region bounded by the curve $y = x^2$ and the line $y = 10$.

Answer

By the symmetry of the parabola, the required area is twice the area of the region $O A B C O$ Now, the area of the region $O A B C O$
$ =\int_0^{10} x d y \text {, where } x^2=y \text { i.e. } x=\sqrt{y}$
$ =\int_0^{10} \sqrt{y} d y=\left[\frac{y^{\frac{3}{2}}}{3 / 2}\right]_0^{10}=\frac{10^{\frac{3}{2}}}{3 / 2}-0=\frac{2 \times 10 \sqrt{10}}{3}=\frac{20 \sqrt{10}}{3}$
$\therefore \text { required area }=2\left[\frac{20 \sqrt{10}}{3}\right]$
$ =\frac{40 \sqrt{10}}{3} \text { sq units. }$

View full question & answer→

Question 33 Marks

Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $x = 3$.

Answer

$\begin{aligned} & \text { Required area }=\text { area of the region OABO } \\ & =2 \text { (area of the region OACO) } \\ & =2 \int_0^3 y d x \text {, where } y^2=4 x \text {, i.e. } y=2 \sqrt{x} \\ & =2 \int_0^3 2 \sqrt{x} d x \\ & =4 \int_0^3 x^{\frac{1}{2}} d x \\ & =4 \cdot\left[\frac{x^{\frac{3}{2}}}{3 / 2}\right]_0^3 \\ & =\frac{8}{3}\left[x^{\frac{3}{2}}\right]_0^3 \\ & =\frac{8}{3}(3 \sqrt{3}-0) \\ & =8 \sqrt{3} \text { sq units. }\end{aligned}$

View full question & answer→

Maths (commerce) Solve the Following Question.(3 Marks)

Take a timed test