Question 11 Mark
Find the area of the region bounded by the curve $y=\cos x$ and the $x$-axis when $0 \leq x \leq \pi$.
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Question 21 Mark
Find the area of the region bounded by the curve $y=x^2$ and the lines $x=0$ and $x=2$.
View full question & answer→Question 31 Mark
Find the area bounded by the curve $y=\sin x$, the ordinate $x=\frac{\pi}{2}$ and the $x$-axis.
Answer
View full question & answer→Area of the curve bounded with $x$-axis
$
\begin{aligned}
& =\int_a^b y d x \\
\text { so required area } & =\int_{\frac{\pi}{2}}^\pi \sin x d x \\
& =(-\cos x)_{\frac{\pi}{2}}^\pi=-\cos \pi+\cos \frac{\pi}{2} \\
& =-(-1)+0=1 \text { square unit. }
\end{aligned}
$
$
\begin{aligned}
& =\int_a^b y d x \\
\text { so required area } & =\int_{\frac{\pi}{2}}^\pi \sin x d x \\
& =(-\cos x)_{\frac{\pi}{2}}^\pi=-\cos \pi+\cos \frac{\pi}{2} \\
& =-(-1)+0=1 \text { square unit. }
\end{aligned}
$
Question 41 Mark
In the interval $\left[0, \frac{\pi}{2}\right]$, find the area of the region bounded by the curve $y =\cos x$ and the $x$-axis.
Answer
View full question & answer→The area of the curve bounded with the $x$-axis $=\int_a^b y d x$
$
\begin{aligned}
\text { Hence required area } & =\int_0^{\pi / 2} \cos x d x \\
& =(\sin x)_0^{\frac{\pi}{2}}=\sin \frac{\pi}{2}-\sin 0 \\
& =1-0=1 \text { square units. }
\end{aligned}
$
$
\begin{aligned}
\text { Hence required area } & =\int_0^{\pi / 2} \cos x d x \\
& =(\sin x)_0^{\frac{\pi}{2}}=\sin \frac{\pi}{2}-\sin 0 \\
& =1-0=1 \text { square units. }
\end{aligned}
$
Question 51 Mark
Find the area bounded by the curve $y=x$, the $x$-axis and the ordinates $x=0$ and $x=a$.
View full question & answer→Question 61 Mark
Find the area of the region bounded by the parabola $y=4 x^2$ and the lines $y=1$ and $y=4$.
Answer
View full question & answer→The area of the curve bounded with $y$-axis
$ \therefore \text { Required area }=2 \int_1^4\left(\frac{y}{4}\right)^{\frac{1}{2}} d y$
$(\because \text { Parabola is symmetrical })$
$=\frac{2}{2} \int_1^4 y^{\frac{1}{2}} d y$
$=\left(\frac{y^{3 / 2}}{3 / 2}\right)_1^4=\frac{2}{3}\left((4)^{3 / 2}-(1)^{3 / 2}\right)$
$=\frac{2}{3}(8-1)=\frac{2 \times 7}{3}$
$=\frac{14}{3} \text { square units. Ans. }$
$ \therefore \text { Required area }=2 \int_1^4\left(\frac{y}{4}\right)^{\frac{1}{2}} d y$
$(\because \text { Parabola is symmetrical })$
$=\frac{2}{2} \int_1^4 y^{\frac{1}{2}} d y$
$=\left(\frac{y^{3 / 2}}{3 / 2}\right)_1^4=\frac{2}{3}\left((4)^{3 / 2}-(1)^{3 / 2}\right)$
$=\frac{2}{3}(8-1)=\frac{2 \times 7}{3}$
$=\frac{14}{3} \text { square units. Ans. }$
Question 71 Mark
Find the area of the region bounded by the curve $y=m x, x$-axis and the ordinates $x=0$ and $x=4$.
Answer
View full question & answer→The area of the curve bounded with $x$-axis $=\int_a^b y d x$
$
\begin{aligned}
\text { Therefore required area }
=\int_0^4 m x d x \\
=m \int_0^4 x d x=m\left(\frac{x^2}{2}\right)_0^4 \\
=\frac{m}{2}\left(4^2-0\right)=\frac{16 m}{2} \\
=8 m \text { square units. }
\end{aligned}
$
$
\begin{aligned}
\text { Therefore required area }
=\int_0^4 m x d x \\
=m \int_0^4 x d x=m\left(\frac{x^2}{2}\right)_0^4 \\
=\frac{m}{2}\left(4^2-0\right)=\frac{16 m}{2} \\
=8 m \text { square units. }
\end{aligned}
$