NCERT Class 12 Maths Misc. Ex. Ch. 8 Solution
Q.1. Find the area under the given curves and given lines:
(i) y = x^2, x = 1, x = 2 and x-axis
(ii) y = x^4, x = 1, x = 5 and x-axis(i) ( y=x^2, x=1, x=2 ) and ( x )-axis
Area of the bounded region
[
\begin{aligned}
& =\int_1^2 x^2 d x=\left[\frac{x^3}{3}\right]_1^2 \
& =\frac{2^3}{3}-\frac{1^3}{3} \
& =\frac{8}{3}-\frac{1}{3} \
& =\frac{7}{3} \text { sq.units. }
\end{aligned}
]
NCERT Class 12 Maths Misc. Ex. Ch. 8 Solution Q. 2
- Sketch the graph of ( y=|x+3| ) and evaluate ( \int_{-6}^0|x+3| d x )
- Sketch the graph of
and evaluate 
Therefore, 
NCERT Class 12 Maths Misc. Ex. Ch. 8 Solution Q. 3
Q. 3. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
\textbf{Q.3 Solution:}\[y = \sin x\]Area of the bounded region:\[= 2 \int_{0}^{\pi} \sin x \, dx\]\[= 2 \left[ -\cos x \right]_{0}^{\pi}\]\[= 2 \left( -\cos\pi – (-\cos 0) \right)\]\[= 2 \left( -(-1) – (-1) \right)\]\[= 2(1 + 1) = 4 \text{ sq. units}\]
NCERT Class 12 Maths Misc. Ex. Ch. 8 Solution Q. 4
Q.4. Area bounded by the curve y=x^3, the X-axis, and the ordinates x=-2 and X=1 is
(a) -9 (b) -15/4 (c) 15/4 (d) 17/4
\[ \text{Soln} \]\[ A = \int_{-2}^{0} y\,dx + \int_{0}^{3} y\,dx \]\[ = \int_{-2}^{0} x^{3}\,dx + \int_{0}^{3} x^{3}\,dx \]\[ = \left[ \frac{x^{4}}{4} \right]_{-2}^{0} + \left[ \frac{x^{4}}{4} \right]_{0}^{3} \]\[ = [0 – 4] + \frac{1}{4} \]\[ = 4 + \frac{1}{4} \]\[ = \frac{17}{4} \text{ sq. units.} \]
NCERT Class 12 Maths Misc. Ex. Ch. 8 Solution Q. 5
\textbf{Q.5:} The area bounded by the curve \[y = x|x|, \quad \text{the x-axis, and the ordinates } x = -1 \text{ and } x = 1\]is given by:\[\text{(a) } 0 \qquad\text{(b) } \frac{1}{3} \qquad\text{(c) } \frac{2}{3} \qquad\text{(d) } \frac{4}{3}\]
\textbf{Solution}\[y = x|x|\]\[|x| = \begin{cases} -x, & x < 0, \\ x, & x \ge 0\end{cases}\]For \(x < 0\):\[y = -x^2\]For \(x \ge 0\):\[y = x^2\]\[\text{Area} = -\int_{-1}^{0} x^2\, dx \;+\; \int_{0}^{1} x^2\, dx\]\[= -\left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1}\]\[= -\left( 0 – (-\tfrac{1}{3}) \right) + \frac{1}{3}\]\[= \frac{1}{3} + \frac{1}{3}= \frac{2}{3}\ \text{sq. units}\]
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