Definite Integral : Area Between Curves

Integration နှင့် ပတ်သက်ပြီး အောက်ပါ သင်ခန်းစာ များကို ရေးသားခဲ့ပါသည်။ ယခုသင်ခန်းစာကို ဖော်ပြပါသင်ခန်းစာများနှင့် တွဲဖက်လေ့လာရမည် ဖြစ်ပါသည်။

1. AREA UNDER A CURVE
2. ANTI-DERIVATIVE
3. PRACTICE PROBLEMS FOR INDEFINITE INTEGRATION
4. INTEGRATION OF $\dfrac{1}{x}$ AND $\dfrac{1}{ax+b}$ EXPONENTIAL FUNCTION
5. INTEGRATION BY SUBSTITUTION
6. INDEFINITE INTEGRATION

Problems

1. The diagram shows parts of the line $y=3 x+10$ and the curve $y=x^{3}-5 x^{2}+3 x+10$. The line and the curve both pass through the point $A$ on the $y$ -axis. The curve has a maximum at the point $B$ and a minimum at the point $C$. The line through $C$, parallel to the $y$ -axis, intersects the line $y=3 x+10$ at the point $D$.
   • Show that the line $A D$ is a tangent to the curve at $A$.
   • Find the $x$ -coordinate of $B$ and of $C$.
   • Find the area of the shaded region $A B C D$, showing all your working.





Line: $y=3 x+10$ When the line cuts $y$ -axis, $x=0$ $\therefore y=10$ Hence, the coordinate of the point $A$ is $(0,10)$ Curve: $y=x^{3}-5 x^{2}+3 x+10$ $y^{\prime}=3 x^{2}-10 x+3$ At $x=0, y^{\prime}=3$ The equation of tangent to the curve at $(0,10)$ is $y-10=3 x$ $y=3 x+10$ At turning points, $y^{\prime}=0$ $\therefore \quad 3 x^{2}-10 x+3=0$ $(3 x-1)(x-3)=0$ When $x=\dfrac{1}{3}, y=\dfrac{283}{27}$ When $x=3, y=1$ $y^{\prime \prime}=9 x-10$ When $x=\dfrac{1}{3}, y^{\prime \prime}=-7<0$ When $x=3, y^{\prime \prime}=17>0$ $\therefore \quad B=\left(\dfrac{1}{3}, \dfrac{283}{27}\right)$ and $C=(3,1)$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{0}^{3}\left[3 x+10-\left(x^{3}-5 x^{2}+3 x+10\right)\right]\ d x$ $=\displaystyle\int_{0}^{3}\left(-x^{3}+5 x^{2}\right)\ d x$ $=\left[-\dfrac{x^{4}}{4}+\dfrac{5 x^{3}}{3}\right]_{0}^{3}$ $=45-\dfrac{8-1}{4}$ $=\dfrac{180-81}{4}$ $=\dfrac{99}{4}$




2. The graph of $y=x^{2}-6 x+10$ cuts the $y$ -axis at $A$. The graphs of $y=x^{2}-6 x+10$ and $y=x+10$ cut one another at $A$ and $B$. The line $B C$ is perpendicular to the $x$ -axis. Calculate the area of the shaded region enclosed by the curve and the line $A B$, showing all your working.





Line: $y_{1}=x+10$ Curve: $y=x^{2}-6 x+10$ When the line intersect the curves, $y_{1}=y_{2}$ $\therefore\ x^{2}-6 x+10=x+10$ $x^{2}-7 x=0$ $x(x-7)=0$ $x=0$ or $x=7$ When $x=0, y=10$ When $x=7, y=17$ $\therefore A=(0,10)$ and $B=(7,17) .$ $\quad$ Area of shaded region $=\displaystyle\int_{0}^{7}\left(y_{1}-y_{2}\right)\ d x$ $=\displaystyle\int_{0}^{7}\left(7 x-x^{2}\right)\ d x$ $=\left[\dfrac{7 x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{0}^{7}$ $=\dfrac{7^{3}(3-2)}{6}=\dfrac{343}{6}$




3. The diagram shows the graph of $y=\sqrt{4+x}$, which meets the $y$ -axis at the point $A$ and the line $x=5$ at the point $B$. Find the area of the region enclosed by the curve and the straight line $A B$.





Curve: $y_{1}=\sqrt{4+x}$ When $x=0, \quad y=\sqrt{4}=2$ When $x=5, \quad y=\sqrt{9}=3$ $\therefore A=(0,2)$ and $B=(5,3)$ $\therefore$ Equation of $A B$ is $y_{2}=\dfrac{3-2}{5-0} x+2$ $\quad =\dfrac{x}{5}+2$ $\begin{aligned} & \text { Area of required region } \\\\ =& \displaystyle\int_{0}^{5}\left(y_{1}-y_{2}\right)\ d x \\\\ =& \displaystyle\int_{0}^{5}\left(\sqrt{4+x}-\dfrac{x}{5}-2\right)\ d x \\\\ =& \displaystyle\int_{0}^{5}(4+x)^{1/2}\ d x-\displaystyle\int_{0}^{5} \dfrac{x}{5} d x-\displaystyle\int_{0}^{5} 2\ d x\\\\ =& \displaystyle\int_{0}^{5}(4+x)^{1 / 2}\ d(4+x)-\displaystyle\int_{0}^{5} \dfrac{x}{5}\ d x-\displaystyle\int_{0}^{5} 2\ d x[\because d(4+x)=d x] \\\\ =& \left[\dfrac{2}{3}(4+x)^{3 / 2}\right]_{0}^{5}-\left[\dfrac{x^{2}}{10}\right]_{0}^{5}-[2 x]_{0}^{5} \\\\ =& \dfrac{2}{3}\left[9^{3 / 2}-4^{3 / 2}\right]-\dfrac{25}{10}-10 \\\\ =& \dfrac{2}{3}[27-8]-\dfrac{5}{2}-10 \\\\ =& \dfrac{38}{3}-\dfrac{25}{2}=\dfrac{1}{6} \end{aligned}$




4. The diagram shows part of the graph of $y=1-2 \cos 3 x$, which crosses the $x$ -axis at the point $A$ and has a maximum at the point $B$.
   • Find the coordinates of $A$.
   • Find the coordinates of $B$.
   • Showing all your working, find the area of the shaded region bounded by the curve, the $x$ -axis and the perpendicular from $B$ to the $x$ -axis.





$y=1-2 \cos 3 x$ When $y=0,$ $1-2 \cos 3 x=0$ $\cos 3 x=\dfrac{1}{2}$ By the diagram, $3 x=\dfrac{\pi}{3}$ $x=\dfrac{\pi}{9}$ $\therefore \quad A=\left(\dfrac{\pi}{9}, 0\right)$ Since $-1 \le \cos 3 x \le 1$ The maximum value of $y=3$, when $\cos 3 x=-1$ By the diagran, $3 x=\pi$ $x=\dfrac{\pi}{3}$ $\therefore B=\left(\dfrac{\pi}{3}, 3\right)$ $\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}}(1-2 \cos 3 x) \ d x \\\\ =& \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}} 1 \ d x-\dfrac{2}{3} \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}} \cos 3 x\ d(3 x) \\\\ =&[x]_{\frac{\pi}{9}}^{\frac{\pi}{3}}-\left[\dfrac{2}{3} \sin 3 x\right]_{\frac{\pi}{9}}^{\frac{\pi}{3}}\\\\ =&\left(\dfrac{\pi}{3}-\dfrac{\pi}{9}\right)-\dfrac{2}{3}\left[\sin \pi-\sin \dfrac{\pi}{3}\right] \\\\ =&\dfrac{2 \pi}{9}-\dfrac{2}{3}\left[0-\dfrac{\sqrt{3}}{2}\right] \\\\ =&\dfrac{2 \pi}{9}+\dfrac{\sqrt{3}}{3} \end{aligned}$




5. The diagram shows part of the curve $y=3 x-x^{\frac{3}{2}}$ and the lines $y=3 x$ and $2 y=27-3 x$. The curve and the line $y=3 x$ meet the $x$ -axis at $O$ and the curve and the line $2 y=27-3 x$ meet the $x$ -axis at $A$.
   • Find the coordinates of $A$.
   • Verify that the coordinates of $B$ are $(3,9)$
   • Find the area of the shaded region.





Line1: $y_{1}=3 x$ Line2: $2 y_{2}=27-3 x$ When line 2 cuts $x$ -axis, $\quad\quad y_{2}=0$ $\therefore\ 27-3 x=0$ $\hspace{1.7cm} x=3 $ $\therefore\ A=(9,0)$ At the point of intersection of Line 1 and line 2, $\quad\quad y_{1}=y_{2}$ $\therefore\ 2(3 x)=27-3 x$ $\quad\quad 9 x=27$ $\quad\quad x=3$ $\therefore\ y=3(3)=9$ $\therefore\ B=(3,9)$ $\quad\quad$ Area of shaded region $\begin{aligned} &= \text { area of } \triangle O A B - \text { area under curve } \\\\ &= \dfrac{1}{2}(9)(9)-\displaystyle\int_{0}^{9}\left(3 x-x^{\frac{3}{2}}\right)\ d x \\\\ &= \dfrac{81}{2}-\left[\dfrac{3 x^{2}}{2}-\dfrac{2 x^{5 / 2}}{5}\right]_{0}^{9} \\\\ &= \dfrac{81}{2}-\dfrac{3(81)}{2}+\frac{2(243)}{5} \\\\ &= \dfrac{81}{5} \end{aligned}$




6. The diagram shows part of the graph of $y=2 \cos \left(x-\dfrac{\pi}{6}\right)$. The graph intersects the $y$ -axis at the point $A$, has a maximum point at $B$ and intersects the $x$ -axis at the point $C$.
   • Find the coordinates of $A$.
   • Find the coordinates of $B$.
   • Find the coordinates of $C$.
   • Find $\displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right) \mathrm{d} x$.
   • Hence find the area of the shaded region.





$y=2 \cos \left(x-\dfrac{\pi}{6}\right)$ When $x=0$, $y=2 \cos \left(-\dfrac{\pi}{6}\right)$ $=\sqrt{3}$ $\therefore\ A=(0, \sqrt{3})$ The maximum value of $y=2 \cos \left(x-\dfrac{\pi}{6}\right)$ is $2$ when $\cos \left(x-\dfrac{\pi}{6}\right)=1$ $x-\dfrac{\pi}{6}=0$ $x=\dfrac{\pi}{6}$ $B=\left(\dfrac{\pi}{6}, 0\right)$ When $y=0,$ $2 \cos \left(x-\dfrac{\pi}{6}\right)=0$ $\cos \left(x-\dfrac{\pi}{6}\right)=0$ $x-\dfrac{\pi}{6}=\dfrac{\pi}{2}$ $x=\dfrac{\pi}{2}+\dfrac{\pi}{6}$ $\therefore x=\dfrac{2 \pi}{3}$ $\therefore C=\left(\dfrac{2 \pi}{3}, 0\right)$ $\displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x$ let $u=x-\dfrac{\pi}{6}$ $\therefore\ d u=d x$ $\begin{aligned} \displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x =&\displaystyle\int 2 \cos u\ d u \\\\ =&2 \sin u+C, c=\text { constant } \\\\ =&2 \sin \left(x-\dfrac{\pi}{6}\right)+C \end{aligned}$ $\begin{aligned} & \text { Area of shaded region } \\\\ =& \displaystyle\int_{0}^{\dfrac{2 \pi}{3}} 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x \\\\ =&\left[2 \sin \left(x-\dfrac{\pi}{6}\right)\right]_{0}^{\frac{2 \pi}{3}} \\\\ =& 2\left[\sin \dfrac{\pi}{2}-\sin \left(-\dfrac{\pi}{6}\right)\right]\\\\ =& 2\left[\sin \dfrac{\pi}{2}+\sin \dfrac{\pi}{6}\right] \\\\ =& 2\left[1+\dfrac{1}{2}\right]= 3 \\\\ \end{aligned}$




7. The diagram shows part of the curve $y=5+4 \tan ^{2}\left(\dfrac{x}{3}\right)$. Hence, find the exact area of the shaded region enclosed by the curve, the $x$ -axis and the lines $x=\dfrac{\pi}{2}$ and $x=\pi$





$\begin{aligned} y&=5+4 \tan ^{2}\left(\dfrac{x}{3}\right) \\\\ &=5+4\left(\sec ^{2}\left(\dfrac{x}{3}\right)-1\right) \\\\ &=4 \sec ^{2}\left(\dfrac{x}{3}\right)+1 \end{aligned}$ $\begin{aligned} &\text { Area of shaded region } \\\\ = & \displaystyle\int_{\dfrac{\pi}{2}}^{\pi}\left[5+4 \tan ^{2}\left(\dfrac{x}{3}\right)\right]\ d x \\\\ = & \displaystyle\int_{\dfrac{\pi}{2}}^{\pi}\left(4 \sec ^{2}\left(\dfrac{x}{3}\right)+1\right)\ d x\\\\ = & 4 \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \sec ^{2}\left(\dfrac{x}{3}\right)\ d x+\displaystyle\int_{\frac{\pi}{2}}^{\pi}1\ d x \\\\ = & \left[\dfrac{4}{3} \tan \left(\dfrac{x}{3}\right)\right]_{\frac{\pi}{2}}^{\pi}+\left[x\right]_{\frac{\pi}{2}}^{\pi} \\\\ = & \frac{4}{3}\left[\tan \dfrac{\pi}{3}-\tan \dfrac{\pi}{6}\right]+\left[\pi-\dfrac{\pi}{2}\right] \\\\ = & 4\left[\sqrt{3}-\dfrac{\sqrt{3}}{3}\right]+\dfrac{\pi}{2} \\\\ = & 8 \sqrt{3}+\dfrac{\pi}{2} \end{aligned}$




8. The diagram shows part of the curve $y=x^{3}+4 x^{2}-5 x+5$ and the line $y=5$. The curve and the line intersect at the points $A, B$ and $C .$ The points $D$ and $E$ are on the $x$ -axis and the lines $A E$ and $C D$ are parallel to the $y$ -axis.Calculate the total area of the shaded regions enclosed between the line and the curve. You must show all your working.





$y=x^{3}+4 x^{2}-5 x+5, y_{2}=5$ When $y=y_{2}$ $x^{3}+4 x^{2}-5 x+5=5$ $x^{3}+4 x^{2}-5 x=0$ $x\left(x^{2}+4 x-5\right)=0$ $\therefore x(x+5)(x-1)=0$ $\therefore x=-5$ or $x=0$ or $x=7$ $\quad\quad$ area of shaded region $\begin{aligned} &=\displaystyle\int_{-5}^{0}\left(y_{1}-y_{2}\right)\ d x+\displaystyle\int_{0}^{1}\left(y_{2}-y_{1}\right)\ d x\\\\ &=\displaystyle\int_{-5}^{0}\left(x^{3}+4 x^{2}-5 x\right)\ d x+\int_{0}^{1}\left(-x^{3}-4 x^{2}+5 x\right)\ d x \\\\\\ &=\left[\dfrac{1}{4} x^{4}+\dfrac{4}{3} x^{3}-\dfrac{5}{2} x^{2}\right]_{-5}^{0}+\left[-\dfrac{1}{4} x^{3}-\dfrac{4}{3} x+\dfrac{5}{2} x^{2}\right]_{0}^{1} \\\\ &=-\left[\dfrac{1}{4}(625)+\dfrac{4}{3}(-125)-\dfrac{5}{2}(25)\right]+\left[-\dfrac{1}{4}-\dfrac{4}{3}+\dfrac{5}{2}\right] \\\\ &=\dfrac{443}{6} \end{aligned}$




9. The diagram shows the curve $y=3 x^{2}-2 x+1 $ and the straight line $y=2 x+5$ intersecting at the points $P$ and $Q$. Showing all your working, find the area of the shaded region.





$y_{1}=2 x+5$ $y_{2}=3 x^{2}-2 x+1$ When $y_{1}=y_{2}$, $2 x+5=3 x^{2}-2 x+1$ $\therefore\ 3 x^{2}-4 x-4=0$ $(3 x+2)(x-2)=0$ $x=-\dfrac{2}{3}$ or $x=2$ When $x=-\dfrac{2}{3}, y=-\dfrac{4}{3}+5=\dfrac{11}{3}$ When $x=2, y=4+5=9$ $\therefore P=\left(-\dfrac{2}{3}, \dfrac{11}{3}\right)$ and $Q=(2,9)$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{-2 / 3}^{2}\left(y_{1}-y_{2}\right)\ d x$ $=\displaystyle\int_{-2 / 3}^{2}\left(-3 x^{2}+4 x+4\right)\ d x$ $=\left[-x^{3}+2 x^{2}+4 x\right]_{-2 / 3}^{2}$ $=-8+8+8-\left(\dfrac{8}{27}+\dfrac{8}{9}-\dfrac{8}{3}\right)$ $=\dfrac{256}{27}$




10. The figure shows part of the curve $y=3 x(x-2) .$ Find the total area enclosed by the curve $y=3 x(x-2)$, the $x$ -axis and the line $x=3$





$y=3 x(x-2)=3 x^{2}-6 x$ When $y=0$, $3 x(x-2)=0$ $\therefore x=0$ or $x=2$ The curve $y=3 x(x-2)$ cuts $x$ -axis at $(0,0)$ and $(2,0)$ $\quad\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{0}^{2}-y d x+\int_{2}^{3} y d x \\\\ =& \displaystyle\int_{0}^{2}\left(-3 x^{2}+6 x\right) d x+\displaystyle\int_{2}^{3}\left(3 x^{2}-6 x\right) d x \\\\ =&\left[-x^{3}+3 x^{2}\right]_{0}^{2}+\left[x^{3}-3 x^{2}\right]_{2}^{3} \\\\ =&[-8+12]+[27-27-8+12] \\\\ =& 8 \end{aligned}$




11. The diagram shows the curve $y=4+3 x-x^{2}$ intersecting the positive $x$ -axis at the point $A$. The line $y=m x+8$ is a tangent to the curve at the point $B$. Find
    • the coordinates of $A$,
    • the value of $m$
    • the coordinates of $B$,
    • the area of the shaded region, showing all your working.





Curve: $y_{1}=4+3 x-x^{2}$ Line: $y_{2}=m x+8$ When the curve ents $x$ -axis, $y=0$ $\therefore\ 4+3 x-x^{2}=0$ $(1+x)(4-x)=0$ $\therefore\ x=-1$ or $x=4$ $\therefore\ A=(4,0)$ Gradient of tangent to the curve $=y_{1}^{\prime}=3-2 x$ $\therefore m=3-2 x$ At the point of contact, $y_{1}=y_{2}$ $\therefore 4+3 x-x^{2}=(3-2 x) x+8$ $x^{2}-4=0$ $x^{2}=4$ $x=\pm 2$ By the diagran, $\quad\quad x=2$ $\therefore\ m=3-2(2)=-1$ $y_{2}=8-x$ When $x=2$ $y=-1(2)+8$ $\quad =6$ $\therefore\ B=(2,6)$ When the tangent cuts $x$ -axis, $8-x=0 \Rightarrow x=8$. $\quad\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{2}^{4}\left(y_{2}-y_{1}\right)\ d x+\int_{4}^{8} y_{2}\ d x \\\\ =& \displaystyle\int_{2}^{4}\left(x^{2}-4 x+4\right)\ d x+\int_{4}^{8}(8-x)\ d x \\\\ =&\left[\dfrac{1}{3} x^{3}-2 x^{2}+4 x\right]_{2}^{4}+\left[8 x-\dfrac{1}{2} x^{2}\right]_{4}^{8} \\\\ =&\left[\dfrac{64}{3}-32+16-\dfrac{8}{3}+8-8\right]+[64-32-32+8] \\\\ =& \dfrac{32}{3} \end{aligned}$




12. The diagram shows part of the curve $y=\dfrac{1}{x+1}$. The shaded region $R$ is bounded by the curve and by the lines $x=1, y=0$ and $x=p$.
    • Find, in terms of $p$, the area of $R$.
    • Hence find, correct to 1 decimal place, the value of $p$ for which the area of $R$ is equal to 2 .





$\quad\quad$ Area of $R$ $= \displaystyle\int_{1}^{p} \dfrac{1}{x+1}\ d x$ $=[\ln |x+1|]_{1}^{p}$ $= \ln |p+1|-\ln 2$ By the problem, $\ln |p+1|-\ln 2=2$ $\ln |p+1|=2+\ln 2$ $p+1=e^{2+\ln 2}$ $\therefore\ p =2 e^{2}-1=13.8$




13. The diagram, which is not drawn to scale, shows part of the graph of $y=8-\mathrm{e}^{2 x}$, crossing the $y$ -axis at $A$. The tangent to the curve at $A$ crosses the $x$ -axis at $B$. Find the area of the shaded region bounded by the curve, the tangent and the $x$ -axis.





Curve: $y=8-e^{2 x}$ When $x=0$ $y=8-e^{0}=7$ $\therefore\ A=(0,7)$ Gradient of tangent to the curve is $y^{\prime}=-2 e^{2 x}$ When $x=0, y^{\prime}=-2$ $\therefore$ Equation of targent at $A$ is $y=-2 x+7$ When the tangent cuts $x$ -axis, $\quad y=0$ $-2 x+7=0$ $x=\dfrac{7}{2}$ $\therefore B=\left(\dfrac{7}{2}, 0\right)$ $8-e^{2 x}=0$ $e^{2 x}=8$ $2 x=\ln 8$ $x=\dfrac{\ln 8}{2}$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{0}^{\frac{\ln 8}{2}}\left(-2 x+7-8+e^{2 x}\right)\ d x+\displaystyle\int_{\frac{\ln 8}{2}}^{7 / 2}(-2 x+7)\ d x$ $=\left[-x^{2}-x+\dfrac{1}{2} e^{2 x}\right]_{0}^{\frac{\ln 8}{2}}+\left[-x^{2}+7 x\right]_{\frac{\ln 8}{2}}^{7 / 2}$ $=-\left(\dfrac{\ln 8}{2}\right)^{2}-\dfrac{\ln 8}{2}+4-\dfrac{1}{2}-\dfrac{49}{4}+\dfrac{49}{2}+\left(\dfrac{\ln 8}{2}\right)^{2}-\dfrac{7}{2} \ln 8$ $=7.4322$




14. The diagram shows the curve $y=x^{3}-3 x^{2}-9 x+k$, where $k$ is a constant. The curve has a minimum point on the $x$ -axis.
    • Find the value of $k$.
    • Find the area of the shaded region.





$y=x^{3}-3 x^{2}-9 x+k$ Let the mimiun porint be $(a, 0)$ $\therefore a^{3}-3 a^{2}-9 a+k=0$ $y^{\prime}=3 u^{2}-6 x-9$ At $(a, 0), y=0$ $\therefore 3 a^{2}-6 a-9=0$ $a^{2}-2 a-3=0$ $(a+1)(a-3)=0$ $\therefore a=-1$ or $a=3$ By the diagran, $a=3$ $\therefore$ minimun ponnt $=(3,0)$ When $a=3, k=27$. $\therefore y=x^{3}-3 x^{2}-9 x+27 .$ $\quad\quad$Area of shaded region $=\displaystyle\int_{0}^{3}\left(x^{3}-3 x^{2}-9 x+27\right)\ d x$ $=\left[\dfrac{x^{4}}{4}-x^{3}-\dfrac{9}{2} x^{2}+27 x\right]_{0}^{3}=\dfrac{135}{4}$




15. The diagram shows the curve $y=x(x-1)(x-2)$, which crosses the $x$ -axis at the points $O(0,0)$, $A(1,0)$ and $B(2,0)$.
    • The tangents to the curve at the points $A$ and $B$ meet at the point $C$. Find the $x$ -coordinate of $C$.
    • Show by integration that the area of the shaded region $R_{1}$ is the same as the area of the shaded region $R_{2}$





Curve: $y=x(x-1)(x-2)$ $\hspace{1cm}y=x^{3}-3 x^{2}+2 x$ $\therefore y^{\prime}=3 x^{2}-6 x+2$ At $x=1, \quad y^{\prime}=-1$ At $x=2, \quad y^{\prime}=2$ Equation of tangent $A$ is $y-0=-1(x-1)$ $y=1-x$ Equation of tangent at $B$ is $y-0=2(x-2)$ $y=2 x-4$ Whent the two tangent intersect each other, $1-x=2 x-4 $ $ x=\dfrac{5}{3}$ When $ x=\dfrac{5}{3}$, $y=1-\dfrac{5}{3}=-\dfrac{2}{3}$ $\therefore\ C=\left(\dfrac{5}{3},-\dfrac{2}{3}\right)$ $\begin{aligned} R_{1} &=\displaystyle\int_{0}^{1}\left(x^{3}-3 x^{2}+2 x\right)\ d x \\\\ &=\left[\dfrac{x^{4}}{4}-x^{3}+x^{2}\right]_{0}^{1} \\\\ &=\dfrac{1}{4}-1+1 \\\\ &=\dfrac{1}{4}\\\\ R_{2} &=\displaystyle\int_{7}^{2}\left(x^{3}-3 x^{2}+2 x\right)\ d x \\\\ &=\left[\dfrac{x^{4}}{4}-x^{3}+x^{2}\right]_{1}^{2} \\\\ &=4-8+4 \dfrac{1}{4}-1+1 \\\\ &=\dfrac{1}{4} \\\\ & \therefore\ R_{1}=R_{2} \end{aligned}$