3 Marks

Question 13 Marks

Find the area of the region bounded by the curve $y=\sin 2 x+\cos 2 x$ and $x=0$ and $x=\frac{\pi}{4}$.

Answer

Where $0 \leq x \leq \frac{\pi}{4} \Rightarrow 0 \leq 2 x \leq \frac{\pi}{2}$ and
$
\begin{array}{r}
\sin 2 x \geq 0 \\
\cos 2 x \geq 0
\end{array}
$
$\because x \in\left(0, \frac{\pi}{4}\right)$ Therefore $y=f(x)=\sin 2 x+\cos 2 x \geq 0$
i.e., the sign of $f(x)$ does not change in the given interval.
Hence required area
$
\begin{array}{l}
=\int_0^{\frac{\pi}{4}} y d x \\
=\int_0^{\frac{\pi}{4}}(\sin 2 x+\cos 2 x) d x \\
=\left(\frac{-\cos 2 x}{2}+\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{4}} \\
=\frac{-1}{2}(\cos 2 x)_0^{\frac{\pi}{4}}+\frac{1}{2}(\sin 2 x)_0^{\frac{\pi}{4}} \\
=\frac{-1}{2}\left(\cos \frac{\pi}{2}-\cos 0\right)+\frac{1}{2}\left(\sin \frac{\pi}{2}-\sin 0\right) \\
=\frac{-1}{2}(0-1)+\frac{1}{2}(1-0) \\
=\frac{1}{2}+\frac{1}{2}=1 \text { square units. }
\end{array}
$

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Question 23 Marks

Find the area enclosed by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Draw the figure.

Answer

The ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ can also be written as follows :
$
\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1
$
In the figure given below, area of the region $ABA ^{\prime} B ^{\prime} A$ enclosed by the ellipse $=4$ (Area of the region AOBA enclosed by the given curve, $x$-axis, ordinates $x=0$, $x=5$ in the first quadrant) (since the ellipse is symmetrical about $x$-axis and $y$-axis both)
$
=4 \int_0^5 y d x
$
Here vertical strips have been taken.

Now we shall find the value of $y$ from the given equation of the ellipse.
$
\begin{array}{l}
\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1 \Rightarrow \frac{y^2}{(4)^2}=1-\frac{x^2}{(5)^2} \\
\Rightarrow \quad y^2=\frac{(4)^2}{(5)^2}\left((5)^2-x^2\right) \\
\Rightarrow \quad y= \pm \frac{4}{5} \sqrt{(5)^2-x^2}
\end{array}
$
Here we shall take only positive value of $y$ since the region $A O B A$ is in the first equadrant. Hence required area
$
\begin{array}{l}
=4 \int_0^5 \frac{4}{5} \sqrt{(5)^2-x^2} d x \\
=\frac{16}{5} \int_0^5 \sqrt{(5)^2-x^2} d x \\
=\frac{16}{5}\left(\frac{x}{5} \sqrt{(5)^2-x^2}+\frac{(5)^2}{2} \sin ^{-1} \frac{x}{5}\right)_0^5 \\
\begin{aligned}
\because \int \sqrt{a^2-x^2} d x & =\frac{x}{a} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a} \\
& =\frac{16}{5}\left(0+\frac{(5)^2}{2} \sin ^{-1} 1-0-0\right) \\
& =\frac{16}{5}\left(\frac{25}{2} \times \frac{\pi}{2}\right) \\
& =20 \pi \text { square units. }
\end{aligned}
\end{array}
$

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