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NCERT Exemplar · Class 12 Mathematics Application of Integrals

34 questions · 34 still being checked

EXERCISE 8.3 1–10 (part 1 of 4)

  1. Exercise 1

    Find the area of the region bounded by the curves y2=9x,y=3x\displaystyle y^2=9 x, y=3 x.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\) sq.units
    The curves meet where\[9x^2 = 9x \Rightarrow x = 0,\ 1 \]\[\text{points } (0,0),\ (1,3) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q1Between them the parabola lies above the line.\[A = \int_0^1 \left(3\sqrt{x} - 3x\right)dx \]\[A = \left[2x^{3/2} - \frac{3x^2}{2}\right]_0^1 \]\[A = 2 - \frac{3}{2} = \frac{1}{2} \]Answer: \(\displaystyle \dfrac{1}{2}\) square unit.

  2. Exercise 2

    Find the area of the region bounded by the parabola y2=2px,x2=2py\displaystyle y^2=2 p x, x^2=2 p y.

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    NCERT’s answer
    \(\displaystyle \frac{4}{3} p^2\) sq. units
    Taking \(\displaystyle p > 0\), the parabolas meet where\[x^2 = 2py \Rightarrow y^2 = \frac{x^4}{4p^2} = 2px \]\[x\left(x^3 - 8p^3\right) = 0 \Rightarrow x = 0,\ 2p \]\[\text{points } (0,0),\ (2p,2p) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q2For \(\displaystyle 0 \le x \le 2p\), \(\displaystyle y^2 = 2px\) lies above \(\displaystyle x^2 = 2py\).\[A = \int_0^{2p}\left(\sqrt{2px} - \frac{x^2}{2p}\right)dx \]\[A = \left[\frac{2\sqrt{2p}}{3}\,x^{3/2} - \frac{x^3}{6p}\right]_0^{2p} \]\[A = \frac{2}{3}(2p)^2 - \frac{8p^3}{6p} = \frac{8p^2}{3} - \frac{4p^2}{3} = \frac{4p^2}{3} \]Answer: \(\displaystyle \dfrac{4p^2}{3}\) square units.

  3. Exercise 3

    Find the area of the region bounded by the curve y=x3\displaystyle y=x^3 and y=x+6\displaystyle y=x+6 and x=0\displaystyle x=0.

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    NCERT’s answer
    $\displaystyle 10$ sq.units
    The cubic and the line meet where\[x^3 = x + 6 \Rightarrow (x-2)\left(x^2 + 2x + 3\right) = 0 \]\[x^2 + 2x + 3 = (x+1)^2 + 2 > 0 \Rightarrow x = 2 \]\[\text{point } (2,8) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q3On \(\displaystyle 0 \le x \le 2\) the line is above the cubic (at \(\displaystyle x=0\): \(\displaystyle 6 > 0\)).\[A = \int_0^2 \left(x + 6 - x^3\right)dx \]\[A = \left[\frac{x^2}{2} + 6x - \frac{x^4}{4}\right]_0^2 \]\[A = 2 + 12 - 4 = 10 \]Answer: \(\displaystyle 10\) square units.

  4. Exercise 4

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

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    NCERT’s answer
    \(\displaystyle \frac{16}{3}\) sq.units
    The parabolas meet where\[x^2 = 4y \Rightarrow y^2 = \frac{x^4}{16} = 4x \]\[x\left(x^3 - 64\right) = 0 \Rightarrow x = 0,\ 4 \]\[\text{points } (0,0),\ (4,4) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q4For \(\displaystyle 0 \le x \le 4\), \(\displaystyle y^2 = 4x\) lies above \(\displaystyle x^2 = 4y\).\[A = \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right)dx \]\[A = \left[\frac{4}{3}x^{3/2} - \frac{x^3}{12}\right]_0^4 \]\[A = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]Answer: \(\displaystyle \dfrac{16}{3}\) square units.

  5. Exercise 5

    Find the area of the region included between y2=9x\displaystyle y^2=9 x and y=x\displaystyle y=x

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    NCERT’s answer
    \(\displaystyle \frac{27}{2}\) sq.units
    The parabola and the line meet where\[x^2 = 9x \Rightarrow x = 0,\ 9 \]\[\text{points } (0,0),\ (9,9) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q5For \(\displaystyle 0 \le x \le 9\), the parabola lies above the line.\[A = \int_0^9 \left(3\sqrt{x} - x\right)dx \]\[A = \left[2x^{3/2} - \frac{x^2}{2}\right]_0^9 \]\[A = 54 - \frac{81}{2} = \frac{27}{2} \]Answer: \(\displaystyle \dfrac{27}{2}\) square units.

  6. Exercise 6

    Find the area of the region enclosed by the parabola x2=y\displaystyle x^2=y and the line y=x+2\displaystyle y=x+2

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    NCERT’s answer
    \(\displaystyle \frac{9}{2}\) sq. units
    The parabola and the line meet where\[x^2 = x + 2 \Rightarrow (x-2)(x+1) = 0 \Rightarrow x = -1,\ 2 \]\[\text{points } (-1,1),\ (2,4) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q6On \(\displaystyle -1 \le x \le 2\) the line is above the parabola.\[A = \int_{-1}^{2} \left(x + 2 - x^2\right)dx \]\[A = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} \]\[A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2} \]Answer: \(\displaystyle \dfrac{9}{2}\) square units.

  7. Exercise 7

    Find the area of region bounded by the line x=2\displaystyle x=2 and the parabola y2=8x\displaystyle y^2=8 x

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    NCERT’s answer
    \(\displaystyle \frac{32}{3}\) sq. units
    The line cuts the parabola where\[y^2 = 8(2) = 16 \Rightarrow y = \pm 4 \]\[\text{points } (2,4),\ (2,-4) \]NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q7The region is symmetric about the \(\displaystyle x\)-axis.\[A = 2\int_0^2 \sqrt{8x}\,dx = 4\sqrt{2}\int_0^2 \sqrt{x}\,dx \]\[A = 4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^2 \]\[A = 4\sqrt{2}\cdot\frac{2}{3}\cdot 2\sqrt{2} = \frac{32}{3} \]Answer: \(\displaystyle \dfrac{32}{3}\) square units.

  8. Exercise 8

    Sketch the region {(x,0):y=4−x2}\displaystyle \left\{(x, 0): y=\sqrt{4-x^2}\right\} and x\displaystyle x-axis. Find the area of the region using integration.

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    NCERT’s answer
    \(\displaystyle 2 \pi\) sq.units
    The curve is the upper half of a circle with centre \(\displaystyle O\) and radius \(\displaystyle 2\).NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q8\[y=\sqrt{4-x^2}\ \Rightarrow\ x^2+y^2=4,\quad y\ge 0 \] \[\text{Area}=\int_{-2}^{2}\sqrt{4-x^2}\,dx=2\int_{0}^{2}\sqrt{4-x^2}\,dx \] \[=2\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac{x}{2}\right]_0^2 \] \[=2\left[\left(0+2\cdot\frac{\pi}{2}\right)-0\right]=2\pi \]Answer: \(\displaystyle 2\pi\) square units.

  9. Exercise 9

    Calcualte the area under the curve y=2x\displaystyle y=2 \sqrt{x} included between the lines x=0\displaystyle x=0 and x=1\displaystyle x=1.

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    NCERT’s answer
    \(\displaystyle \frac{4}{3}\) sq.units
    NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q9\[A=\int_0^1 2\sqrt{x}\,dx=2\int_0^1 x^{1/2}\,dx \] \[=2\left[\frac{x^{3/2}}{3/2}\right]_0^1=\frac43\left[x^{3/2}\right]_0^1 \] \[=\frac43(1-0)=\frac43 \]Answer: \(\displaystyle \dfrac{4}{3}\) square units.

  10. Exercise 10

    Using integration, find the area of the region bounded by the line 2y=5x+7\displaystyle 2 y=5 x+7, x\displaystyle x-axis and the lines x=2\displaystyle x=2 and x=8\displaystyle x=8.

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    NCERT’s answer
    $\displaystyle 96$ sq.units
    NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q10\[2y=5x+7\ \Rightarrow\ y=\frac{5x+7}{2}>0\quad\text{for } 2\le x\le 8 \] \[A=\int_2^8\frac{5x+7}{2}\,dx=\frac12\left[\frac{5x^2}{2}+7x\right]_2^8 \] \[=\frac12\big[(160+56)-(10+14)\big]=\frac12(216-24) \] \[=96 \]Answer: \(\displaystyle 96\) square units.