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NCERT Solutions · Class 12 Mathematics Application of Integrals

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Miscellaneous Exercise 1–5 (part 2 of 2)

  1. Exercise 1

    Find the area under the given curves and given lines:
    (i)
    y=x2,x=1,x=2\displaystyle y=x^{2}, x=1, x=2 and x\displaystyle x-axis
    (ii)
    y=x4,x=1,x=5\displaystyle y=x^{4}, x=1, x=5 and x\displaystyle x-axis

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{7}{3}\) (ii) $\displaystyle 624.8$
    The area of the region bounded by a curve \(\displaystyle y=f(x)\), the \(\displaystyle x\)-axis and the ordinates \(\displaystyle x=a,\;x=b\) is \(\displaystyle \int_{a}^{b}|y|\,dx\); when \(\displaystyle f(x)\ge 0\) throughout \(\displaystyle [a,b]\) the modulus may be dropped and the area is simply \(\displaystyle \int_{a}^{b}y\,dx\).
    (i)
    On \(\displaystyle [1,2]\) we have \(\displaystyle y=x^{2}>0\), so the whole strip lies above the \(\displaystyle x\)-axis:
    \[\text{Area}=\int_{1}^{2}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{1}^{2}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}.\]
    Area \(\displaystyle =\dfrac{7}{3}\) square units.
    (ii)
    On \(\displaystyle [1,5]\) we have \(\displaystyle y=x^{4}>0\), so again no splitting is needed:
    \[\text{Area}=\int_{1}^{5}x^{4}\,dx=\left[\frac{x^{5}}{5}\right]_{1}^{5}=\frac{3125}{5}-\frac{1}{5}=\frac{3124}{5}.\]
    Area \(\displaystyle =\dfrac{3124}{5}=624.8\) square units.

  2. Exercise 2

    Sketch the graph of y=∣x+3∣\displaystyle y=|x+3| and evaluate ∫−60∣x+3∣dx\displaystyle \int_{-6}^{0}|x+3| d x.

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    NCERT’s answer
    $\displaystyle 9$
    NCERT_Solution_Class12_Maths_Ch8_Misc_Q2By the definition of the modulus, \[|x+3|=\begin{cases}-(x+3), & x<-3,\\[2pt] x+3, & x\ge -3.\end{cases}\] So the graph is two straight half-lines meeting at the vertex \(\displaystyle (-3,0)\): the line \(\displaystyle y=-(x+3)\) of slope \(\displaystyle -1\) to the left of \(\displaystyle x=-3\) (through \(\displaystyle (-6,3)\)), and the line \(\displaystyle y=x+3\) of slope \(\displaystyle +1\) to the right (through \(\displaystyle (0,3)\)). The graph never dips below the \(\displaystyle x\)-axis.Because the integrand changes formula at \(\displaystyle x=-3\), the integral must be split there (this is the step to get right \(\displaystyle -\) integrating \(\displaystyle x+3\) straight through from \(\displaystyle -6\) to \(\displaystyle 0\) would give \(\displaystyle 0\)): \[\int_{-6}^{0}|x+3|\,dx=\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{0}(x+3)\,dx.\] First piece: \[\int_{-6}^{-3}-(x+3)\,dx=-\left[\frac{x^{2}}{2}+3x\right]_{-6}^{-3}=-\left[\left(\frac{9}{2}-9\right)-\left(18-18\right)\right]=-\left(-\frac{9}{2}\right)=\frac{9}{2}.\] Second piece: \[\int_{-3}^{0}(x+3)\,dx=\left[\frac{x^{2}}{2}+3x\right]_{-3}^{0}=0-\left(\frac{9}{2}-9\right)=\frac{9}{2}.\] Adding, \[\int_{-6}^{0}|x+3|\,dx=\frac{9}{2}+\frac{9}{2}=9.\] (Check against the sketch: two right triangles of legs \(\displaystyle 3\) and \(\displaystyle 3\), total area \(\displaystyle 2\times\tfrac12\times3\times3=9\).)\(\displaystyle \displaystyle\int_{-6}^{0}|x+3|\,dx=9\).

  3. Exercise 3

    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
    $\displaystyle 4$
    Area is \(\displaystyle \int|y|\,dx\), so the interval must first be split at every point where \(\displaystyle y\) changes sign. Here \(\displaystyle y=\sin x\) satisfies \[\sin x\ge 0 \text{ on } [0,\pi],\qquad \sin x\le 0 \text{ on } [\pi,2\pi],\] so the arch over \(\displaystyle [0,\pi]\) lies above the \(\displaystyle x\)-axis and the arch over \(\displaystyle [\pi,2\pi]\) lies below it. The two areas must be added as positive quantities.Above the axis: \[A_{1}=\int_{0}^{\pi}\sin x\,dx=\big[-\cos x\big]_{0}^{\pi}=-\cos\pi+\cos 0=1+1=2.\] Below the axis: \[\int_{\pi}^{2\pi}\sin x\,dx=\big[-\cos x\big]_{\pi}^{2\pi}=-\cos 2\pi+\cos\pi=-1-1=-2,\qquad A_{2}=|-2|=2.\] Hence \[\text{Area}=A_{1}+A_{2}=2+2=4.\] Note that \(\displaystyle \int_{0}^{2\pi}\sin x\,dx=0\); that zero is the signed area and is not the answer, because the two arches cancel.Required area \(\displaystyle =4\) square units.
  4. Choose the correct answer in the following Exercises from $\displaystyle 4$ to 5.

    Exercise 4

    Area bounded by the curve y=x3\displaystyle y=x^{3}, the x\displaystyle x-axis and the ordinates x=−2\displaystyle x=-2 and x=1\displaystyle x=1 is (A) - 9\displaystyle 9 (B) −154\displaystyle \frac{-15}{4} (C) 154\displaystyle \frac{15}{4} (D) 174\displaystyle \frac{17}{4}

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    NCERT’s answer
    D
    The bounded area is \(\displaystyle \int_{-2}^{1}|y|\,dx\), so split the interval where \(\displaystyle y=x^{3}\) changes sign, namely at \(\displaystyle x=0\): \(\displaystyle x^{3}\le 0\) on \(\displaystyle [-2,0]\) and \(\displaystyle x^{3}\ge 0\) on \(\displaystyle [0,1]\).Portion below the \(\displaystyle x\)-axis: \[\int_{-2}^{0}x^{3}\,dx=\left[\frac{x^{4}}{4}\right]_{-2}^{0}=0-\frac{16}{4}=-4,\qquad A_{1}=|-4|=4.\] Portion above the \(\displaystyle x\)-axis: \[A_{2}=\int_{0}^{1}x^{3}\,dx=\left[\frac{x^{4}}{4}\right]_{0}^{1}=\frac{1}{4}.\] Therefore \[\text{Area}=A_{1}+A_{2}=4+\frac{1}{4}=\frac{17}{4}.\] (Integrating straight through gives \(\displaystyle \tfrac14-4=-\tfrac{15}{4}\), the signed area, which is the trap behind options (A) and (B).)Correct answer: (D) \(\displaystyle \dfrac{17}{4}\).

  5. Exercise 5

    The area bounded by the curve y=x∣x∣,x\displaystyle y=x|x|, x-axis and the ordinates x=−1\displaystyle x=-1 and x=1\displaystyle x=1 is given by (A) 0\displaystyle 0 (B) 13\displaystyle \frac{1}{3} (C) 23\displaystyle \frac{2}{3} (D) 43\displaystyle \frac{4}{3} [Hint : y=x2\displaystyle y=x^{2} if x>0\displaystyle x>0 and y=−x2\displaystyle y=-x^{2} if x<0\displaystyle x<0 ].

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    NCERT’s answer
    C
    Unfold the modulus first, as the hint indicates: \[y=x|x|=\begin{cases}x^{2}, & x\ge 0,\\[2pt] -x^{2}, & x<0.\end{cases}\] So on \(\displaystyle [-1,0]\) the curve lies below the \(\displaystyle x\)-axis and on \(\displaystyle [0,1]\) it lies above it; the area is \(\displaystyle \int_{-1}^{1}|y|\,dx\), split at \(\displaystyle x=0\).Left part (below the axis): \[\int_{-1}^{0}(-x^{2})\,dx=-\left[\frac{x^{3}}{3}\right]_{-1}^{0}=-\left(0+\frac{1}{3}\right)=-\frac{1}{3},\qquad A_{1}=\left|-\frac{1}{3}\right|=\frac{1}{3}.\] Right part (above the axis): \[A_{2}=\int_{0}^{1}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}.\] Hence \[\text{Area}=A_{1}+A_{2}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.\] (Since \(\displaystyle y=x|x|\) is an odd function, \(\displaystyle \int_{-1}^{1}y\,dx=0\); that is option (A) and it is the signed area, not the area.)Correct answer: (C) \(\displaystyle \dfrac{2}{3}\).