Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 8 Miscellaneous Exercise page covers all 5 questions, solved step-by-step — areas under power curves, sketching and integrating an absolute-value graph, the area under a full sine wave, and two MCQs on cubic-type curves.
The area under y=x^2 between x=1 and x=2, above the x-axis, is \text{Area}=\displaystyle\int_{1}^{2} x^2\,dx=\left[\dfrac{x^3}{3}\right]_1^2=\dfrac{8}{3}-\dfrac{1}{3}=\dfrac{7}{3}.
The area under y=x^4 between x=1 and x=5, above the x-axis, is \text{Area}=\displaystyle\int_{1}^{5} x^4\,dx=\left[\dfrac{x^5}{5}\right]_1^5=\dfrac{3125}{5}-\dfrac{1}{5}=\dfrac{3124}{5}.
The graph of y=|x+3| is a V-shape with its vertex at (-3,0).
It has two linear pieces: for x\ge -3, y=x+3 (rising); for x \lt -3, y=-(x+3) (falling).
Splitting the interval [-6,0] at the vertex x=-3:
\displaystyle\int_{-6}^{0}|x+3|\,dx=\displaystyle\int_{-6}^{-3}-(x+3)\,dx+\displaystyle\int_{-3}^{0}(x+3)\,dx.
First piece: \displaystyle\int_{-6}^{-3}-(x+3)\,dx=\left[-\dfrac{x^2}{2}-3x\right]_{-6}^{-3}=\left(-\dfrac{9}{2}+9\right)-\left(-18+18\right)=\dfrac{9}{2}.
Second piece: \displaystyle\int_{-3}^{0}(x+3)\,dx=\left[\dfrac{x^2}{2}+3x\right]_{-3}^{0}=0-\left(\dfrac{9}{2}-9\right)=\dfrac{9}{2}.
Adding both pieces: \dfrac{9}{2}+\dfrac{9}{2}=9.
The sine curve lies above the x-axis on [0,\pi] and below it on [\pi,2\pi], so the required area must be found as the sum of the absolute values of the two integrals, not their direct sum (which would cancel to zero).
\text{Area}=\left|\displaystyle\int_{0}^{\pi}\sin x\,dx\right|+\left|\displaystyle\int_{\pi}^{2\pi}\sin x\,dx\right|.
\displaystyle\int_{0}^{\pi}\sin x\,dx=\left[-\cos x\right]_0^{\pi}=(-\cos\pi)-(-\cos 0)=1+1=2.
\displaystyle\int_{\pi}^{2\pi}\sin x\,dx=\left[-\cos x\right]_{\pi}^{2\pi}=(-\cos 2\pi)-(-\cos\pi)=-1-1=-2, so its absolute value is 2.
Total area: 2+2=4.
Since y=x^3 is negative for x \lt 0 and positive for x \gt 0, split the interval at x=0 and take absolute values separately — a single integral from -2 to 1 would let the negative and positive parts cancel.
\displaystyle\int_{-2}^{0}x^3\,dx=\left[\dfrac{x^4}{4}\right]_{-2}^{0}=0-4=-4, so its absolute value is 4.
\displaystyle\int_{0}^{1}x^3\,dx=\left[\dfrac{x^4}{4}\right]_0^1=\dfrac{1}{4}.
Total area: 4+\dfrac{1}{4}=\dfrac{17}{4}.
By definition, y=x^2 when x \gt 0 and y=-x^2 when x \lt 0. So split the integral at x=0.
\displaystyle\int_{-1}^{0}(-x^2)\,dx=\left[-\dfrac{x^3}{3}\right]_{-1}^{0}=0-\dfrac{1}{3}=-\dfrac{1}{3}, so its absolute value is \dfrac{1}{3}.
\displaystyle\int_{0}^{1}x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{3}.
Total area: \dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}.
You've worked through every exercise in Application of Integrals. Next up is Chapter 9, Differential Equations — a natural continuation from integration.
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