2 Marks Questions

Question 12 Marks

Find the area bounded by the curve y = cos x between x = 0 and x = 2$\pi$

Answer

From the Fig., the required area = area of the region OABO + area of the region BCDB + area of the region DEFD.

Thus, we have the required area
$=\int_{0}^{\frac{\pi}{2}} \cos x d x+\left|\int_{-\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x d x\right|+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x$
$=[\sin x]_{0}^{\frac{\pi}{2}}+\left|[\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}\right|+[\sin x]_{\frac{3 \pi}{2}}^{2 \pi}$
= 1 + 2 + 1 = 4

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Question 22 Marks

Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = - 1 and x = 1

Answer

As shown in the fig., the line y = 3x + 2 meets x - axis A= $\frac{-2}{3}$ and its graph lies below x - axis for $x \in\left(-1, \frac{-2}{3}\right)$ and above x - axis for $x \in\left(\frac{-2}{3}, 1\right)$

The required area = Area of the region ACBA + Area of the region ADEA
$=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^{1}(3 x+2) d x$
$\left.=\big[| \frac{3 x^{2}}{2}+2 x\right]\left._{-1}^{\frac{-2}{3}}\right|_{-1} ^{\frac{2}{3}} |+\left[\frac{3 x^{2}}{2}+2 x\right]_{\frac{-2}{3}}^{1}=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}$

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Question 32 Marks

Find the area of the region bounded by two parabolas $y=x^2$ and $y^2=x$.

Answer

Intersection points: Solve $x^2=\sqrt{x}$. Squaring both sides gives $x^4=x \Longrightarrow x\left(x^3-\right.$ $1)=0$. Points are $(0,0)$ and $(1,1)$.
Setup Integral: Area $A=\int_0^1\left(y_{\text {upper }}-y_{\text {lower }}\right) d x . A=\int_0^1\left(\sqrt{x}-x^2\right) d x$
Integrate: $A=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^3}{3}\right]_0^1=\left[\frac{2}{3} x^{3 / 2}-\frac{1}{3} x^3\right]_0^1$
Evaluate: $\left(\frac{2}{3}-\frac{1}{3}\right)-(0)=\frac{1}{3}$.
The area is $1 / 3$ square units.

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MATHS 2 Marks Questions

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