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NCERT Exemplar · Class 12 Mathematics Application of Integrals

34 questions · 34 still being checked

EXERCISE 8.3 11–20 (part 2 of 4)

  1. Exercise 11

    Draw a rough sketch of the curve y=x−1\displaystyle y=\sqrt{x-1} in the interval [1,5]\displaystyle [1,5]. Find the area under the curve and between the lines x=1\displaystyle x=1 and x=5\displaystyle x=5.

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    NCERT’s answer
    \(\displaystyle \frac{16}{3}\) sq.units
    The curve is the upper half of the parabola \(\displaystyle y^2=x-1\), with vertex \(\displaystyle (1,0)\).\[x=1,\ 2,\ 5\ \Rightarrow\ y=0,\ 1,\ 2 \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q11\[A=\int_1^5\sqrt{x-1}\,dx=\left[\frac23(x-1)^{3/2}\right]_1^5 \] \[=\frac23\left(4^{3/2}-0\right)=\frac23\cdot 8=\frac{16}{3} \]Answer: \(\displaystyle \dfrac{16}{3}\) square units.

  2. Exercise 12

    Determine the area under the curve y=a2−x2\displaystyle y=\sqrt{a^2-x^2} included between the lines x=0\displaystyle x=0 and x=a\displaystyle x=a.

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    NCERT’s answer
    \(\displaystyle \frac{\pi a^2}{4}\) sq. units
    NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q12\[A=\int_0^a\sqrt{a^2-x^2}\,dx=\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a \] \[=\left(0+\frac{a^2}{2}\cdot\frac{\pi}{2}\right)-(0+0)=\frac{\pi a^2}{4} \]Answer: \(\displaystyle \dfrac{\pi a^2}{4}\) square units.

  3. Exercise 13

    Find the area of the region bounded by y=x\displaystyle y=\sqrt{x} and y=x\displaystyle y=x.

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    NCERT’s answer
    \(\displaystyle \frac{1}{6}\) sq. units
    \[\sqrt{x}=x\ \Rightarrow\ x=x^2\ \Rightarrow\ x(x-1)=0\ \Rightarrow\ x=0,\ 1 \]The curves meet at \(\displaystyle (0,0)\) and \(\displaystyle (1,1)\), and \(\displaystyle \sqrt{x}\ge x\) on \(\displaystyle [0,1]\).NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q13\[A=\int_0^1\left(\sqrt{x}-x\right)dx=\left[\frac23x^{3/2}-\frac{x^2}{2}\right]_0^1 \] \[=\frac23-\frac12=\frac16 \]Answer: \(\displaystyle \dfrac{1}{6}\) square unit.

  4. Exercise 14

    Find the area enclosed by the curve y=−x2\displaystyle y=-x^2 and the straight lilne x+y+2=0\displaystyle x+y+2=0.

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    NCERT’s answer
    \(\displaystyle \frac{9}{2}\) sq. units
    \[x+y+2=0\ \Rightarrow\ y=-x-2 \] \[-x^2=-x-2\ \Rightarrow\ x^2-x-2=0\ \Rightarrow\ (x-2)(x+1)=0 \] \[x=-1,\ 2\ \Rightarrow\ \text{points } (-1,-1),\ (2,-4) \]At \(\displaystyle x=0\) the parabola gives \(\displaystyle 0\) and the line \(\displaystyle -2\), so the parabola lies above the line on \(\displaystyle [-1,2]\).NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q14\[A=\int_{-1}^{2}\left[-x^2-(-x-2)\right]dx=\int_{-1}^{2}\left(-x^2+x+2\right)dx \] \[=\left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{-1}^{2} \] \[=\left(-\frac83+2+4\right)-\left(\frac13+\frac12-2\right)=\frac{10}{3}+\frac76=\frac92 \]Answer: \(\displaystyle \dfrac{9}{2}\) square units.

  5. Exercise 15

    Find the area bounded by the curve y=x,x=2y+3\displaystyle y=\sqrt{x}, x=2 y+3 in the first quadrant and x\displaystyle x-axis.

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    NCERT’s answer
    $\displaystyle 9$ sq.units
    The line meets the \(\displaystyle x\)-axis at \(\displaystyle x=3\). It meets the curve \(\displaystyle x=y^2\) where\[y^2 = 2y+3 \] \[(y-3)(y+1)=0 \Rightarrow y=3 \quad (y>0) \] \[\text{Point of intersection: } (9,3) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q15For each \(\displaystyle y\in[0,3]\), \(\displaystyle x\) runs from the curve to the line.\[\text{Area}=\int_0^3\left(2y+3-y^2\right)dy \] \[=\left[y^2+3y-\frac{y^3}{3}\right]_0^3 \] \[=9+9-9=9 \]Answer: \(\displaystyle 9\) square units

  6. Exercise 16

    Find the area of the region bounded by the curve y2=2x\displaystyle y^2=2 x and x2+y2=4x\displaystyle x^2+y^2=4 x.

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    NCERT’s answer
    \(\displaystyle 2\left[\pi-\frac{8}{3}\right]\) sq.units
    The circle has centre \(\displaystyle (2,0)\) and radius \(\displaystyle 2\). The curves meet where\[y^2=2x \ \text{ and } \ x^2+y^2=4x \Rightarrow x^2-2x=0 \] \[x=0 \ \text{ or } \ x=2 \Rightarrow (0,0),\ (2,2),\ (2,-2) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q16For \(\displaystyle 0\le x\le 2\) the circle lies above the parabola, and the region between them is symmetric about the \(\displaystyle x\)-axis.\[\text{Area}=2\int_0^2\left[\sqrt{4x-x^2}-\sqrt{2x}\right]dx \] \[\int_0^2\sqrt{4-(x-2)^2}\,dx=\left[\frac{x-2}{2}\sqrt{4x-x^2}+2\sin^{-1}\frac{x-2}{2}\right]_0^2=\pi \] \[\int_0^2\sqrt{2x}\,dx=\sqrt2\cdot\frac23\,x^{3/2}\Big|_0^2=\frac83 \] \[\text{Area}=2\left(\pi-\frac83\right)=2\pi-\frac{16}{3} \]Answer: \(\displaystyle \dfrac23\,(3\pi-8)\) square units

  7. Exercise 17

    Find the area bounded by the curve y=sin⁡x\displaystyle y=\sin x between x=0\displaystyle x=0 and x=2π\displaystyle x=2 \pi.

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    NCERT’s answer
    $\displaystyle 4$ sq.units
    \(\displaystyle \sin x\ge0\) on \(\displaystyle [0,\pi]\) and \(\displaystyle \sin x\le0\) on \(\displaystyle [\pi,2\pi]\), so the second arc lies below the \(\displaystyle x\)-axis.NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q17\[\text{Area}=\int_0^{\pi}\sin x\,dx+\left|\int_{\pi}^{2\pi}\sin x\,dx\right| \] \[\int_0^{\pi}\sin x\,dx=\Big[-\cos x\Big]_0^{\pi}=1+1=2 \] \[\int_{\pi}^{2\pi}\sin x\,dx=\Big[-\cos x\Big]_{\pi}^{2\pi}=-1-1=-2 \] \[\text{Area}=2+|-2|=4 \]Answer: \(\displaystyle 4\) square units

  8. Exercise 18

    Find the area of region bounded by the triangle whose vertices are (−1,1),(0,5)\displaystyle (-1,1),(0,5) and (3,2)\displaystyle (3,2), using integration.

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    NCERT’s answer
    \(\displaystyle \frac{15}{2}\) sq. units
    Let \(\displaystyle A(-1,1)\), \(\displaystyle B(0,5)\), \(\displaystyle C(3,2)\). The sides are\[AB:\ y-1=\frac{5-1}{0+1}(x+1) \Rightarrow y=4x+5 \] \[BC:\ y-5=\frac{2-5}{3-0}\,x \Rightarrow y=5-x \] \[CA:\ y-1=\frac{2-1}{3+1}(x+1) \Rightarrow y=\frac{x+5}{4} \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q18Above \(\displaystyle CA\), the upper boundary is \(\displaystyle AB\) for \(\displaystyle -1\le x\le0\) and \(\displaystyle BC\) for \(\displaystyle 0\le x\le3\).\[\text{Area}=\int_{-1}^{0}\left[(4x+5)-\frac{x+5}{4}\right]dx+\int_0^3\left[(5-x)-\frac{x+5}{4}\right]dx \] \[=\int_{-1}^{0}\frac{15}{4}(x+1)\,dx+\int_0^3\frac54(3-x)\,dx \] \[=\frac{15}{8}\Big[(x+1)^2\Big]_{-1}^{0}-\frac58\Big[(3-x)^2\Big]_0^3 \] \[=\frac{15}{8}+\frac{45}{8}=\frac{15}{2} \]Answer: \(\displaystyle \dfrac{15}{2}\) square units

  9. Exercise 19

    Draw a rough sketch of the region {(x,y):y2≤6ax\displaystyle \left\{(x, y): y^2 \leq 6 a x\right. and x2+y2≤16a2}\displaystyle \left.x^2+y^2 \leq 16 a^2\right\}. Also find the area of the region sketched using method of integration.

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    Take \(\displaystyle a>0\). The curves \(\displaystyle y^2=6ax\) and \(\displaystyle x^2+y^2=16a^2\) meet where\[x^2+6ax-16a^2=0 \] \[(x+8a)(x-2a)=0 \Rightarrow x=2a,\ y=\pm2\sqrt3\,a \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q19The region lies inside both curves and is symmetric about the \(\displaystyle x\)-axis. The parabola bounds it for \(\displaystyle 0\le x\le 2a\), the circle for \(\displaystyle 2a\le x\le 4a\).\[\text{Area}=2\left[\int_0^{2a}\sqrt{6ax}\,dx+\int_{2a}^{4a}\sqrt{16a^2-x^2}\,dx\right] \] \[\int_0^{2a}\sqrt{6ax}\,dx=\sqrt{6a}\cdot\frac23(2a)^{3/2}=\frac{8\sqrt3}{3}a^2 \] \[\int_{2a}^{4a}\sqrt{16a^2-x^2}\,dx=\left[\frac x2\sqrt{16a^2-x^2}+8a^2\sin^{-1}\frac{x}{4a}\right]_{2a}^{4a}=\frac{8\pi}{3}a^2-2\sqrt3\,a^2 \] \[\text{Area}=2\left[\frac{8\sqrt3}{3}a^2+\frac{8\pi}{3}a^2-2\sqrt3\,a^2\right]=\frac{4a^2}{3}\left(4\pi+\sqrt3\right) \]Answer: \(\displaystyle \dfrac{4a^2}{3}\left(4\pi+\sqrt3\right)\) square units

  10. Exercise 20

    Compute the area bounded by the lines x+2y=2,y−x=1\displaystyle x+2 y=2, y-x=1 and 2x+y=7\displaystyle 2 x+y=7.

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    NCERT’s answer
    $\displaystyle 6$ sq.units
    Pairwise intersections of the lines give the vertices.\[x+2y=2,\ y-x=1 \Rightarrow A(0,1) \] \[y-x=1,\ 2x+y=7 \Rightarrow B(2,3) \] \[x+2y=2,\ 2x+y=7 \Rightarrow C(4,-1) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q20The lower boundary is \(\displaystyle AC:\ y=\dfrac{2-x}{2}\). The upper boundary is \(\displaystyle AB:\ y=x+1\) for \(\displaystyle 0\le x\le2\) and \(\displaystyle BC:\ y=7-2x\) for \(\displaystyle 2\le x\le4\).\[\text{Area}=\int_0^2\left[(x+1)-\frac{2-x}{2}\right]dx+\int_2^4\left[(7-2x)-\frac{2-x}{2}\right]dx \] \[=\int_0^2\frac{3x}{2}\,dx+\int_2^4\left(6-\frac{3x}{2}\right)dx \] \[=\Big[\frac{3x^2}{4}\Big]_0^2+\Big[6x-\frac{3x^2}{4}\Big]_2^4 \] \[=3+3=6 \]Answer: \(\displaystyle 6\) square units