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Exercise 9.8 · Q3

Q.Find the area of the region bounded by the curve 2+x−x2+y=02+x-x^2+y=0, xx-axis, x=−3x=-3 and x=3x=3.

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✓ Free question

Factor the parabola to find its zeros inside [−3,3][-3,3], split into three constant-sign pieces, and add absolute values.

Step 1. Factor the curve and find its zeros. 2+x−x2+y=0⇒y=x2−x−2=(x−2)(x+1)2+x-x^2+y=0\Rightarrow y=x^2-x-2=(x-2)(x+1), zero at x=−1,2x=-1,2; both lie inside [−3,3][-3,3].

Step 2. Determine the sign of yy on each piece. Since the parabola opens upward, y≥0y\ge0 outside the roots and y≤0y\le0 between them: at x=−3x=-3, y=(−5)(−2)=10>0y=(-5)(-2)=10>0; at x=0x=0, y=(−2)(1)=−2<0y=(-2)(1)=-2<0; at x=3x=3, y=(1)(4)=4>0y=(1)(4)=4>0. So

A=∫−3−1y dx+∣∫−12y dx∣+∫23y dx.A=\int_{-3}^{-1}y\,dx+\left|\int_{-1}^{2}y\,dx\right|+\int_{2}^{3}y\,dx.

Step 3. Antiderivative. F(x)=∫(x2−x−2) dx=x33−x22−2x+CF(x)=\displaystyle\int(x^2-x-2)\,dx=\dfrac{x^3}3-\dfrac{x^2}2-2x+C.

Step 4. Evaluate FF at the four endpoints. F(−3)=−9−92+6=−152F(-3)=-9-\dfrac92+6=-\dfrac{15}2; F(−1)=−13−12+2=76F(-1)=-\dfrac13-\dfrac12+2=\dfrac76; F(2)=83−2−4=−103F(2)=\dfrac83-2-4=-\dfrac{10}3; F(3)=9−92−6=−32F(3)=9-\dfrac92-6=-\dfrac32.

Step 5. Combine each piece. ∫−3−1y dx=F(−1)−F(−3)=76+152=263\displaystyle\int_{-3}^{-1}y\,dx=F(-1)-F(-3)=\dfrac76+\dfrac{15}2=\dfrac{26}3.

∫−12y dx=F(2)−F(−1)=−103−76=−92\displaystyle\int_{-1}^{2}y\,dx=F(2)-F(-1)=-\dfrac{10}3-\dfrac76=-\dfrac92, so its absolute value is 92\dfrac92.

∫23y dx=F(3)−F(2)=−32+103=116\displaystyle\int_{2}^{3}y\,dx=F(3)-F(2)=-\dfrac32+\dfrac{10}3=\dfrac{11}6.

Step 6. Add. A=263+92+116=526+276+116=906=15A=\dfrac{26}3+\dfrac92+\dfrac{11}6=\dfrac{52}6+\dfrac{27}6+\dfrac{11}6=\dfrac{90}6=15.

✓Final answer

Area =15=15 sq units.

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