1 Marks

Question 11 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

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}
$

View full question & answer→

Question 21 Mark

Find the area of the region bounded by the curve $y=\cos x$ and the $x$-axis when $0 \leq x \leq \pi$.

View full question & answer→

Question 31 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 41 Mark

Find the area bounded by the curve $y=\sin x$, the ordinate $x=\frac{\pi}{2}$ and the $x$-axis.

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}
$

View full question & answer→

Question 51 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

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}
$

View full question & answer→

Question 61 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 71 Mark

Find the area of the region bounded by the parabola $y=4 x^2$ and the lines $y=1$ and $y=4$.

Answer

The area of the curve bounded with $y$-axis

$\begin{array}{l} \therefore \text { Required area }=2 \int_1^4\left(\frac{y}{4}\right)^{\frac{1}{2}} d y \\ (\because \text { Parabola is symmetrical }) \\ =\frac{2}{2} \int_1^4 y^{\frac{1}{2}} d y \\ =\left(\frac{y^{3 / 2}}{3 / 2}\right)_1^4=\frac{2}{3}\left((4)^{3 / 2}-(1)^{3 / 2}\right) \\ =\frac{2}{3}(8-1)=\frac{2 \times 7}{3} \\ =\frac{14}{3} \text { square units. Ans. }\end{array}$

View full question & answer→

Maths 1 Marks

Test yourself on this topic