# NCERT Solutions Class 12 Maths (Application Of Integrals) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1.

Answer. (i) The required area is represented by the shaded area ADCBA as (ii) The required area is represented by the shaded area ADCBA as Q2. Find the area between the curve

Answer. The required area is represented by the shaded area OBAO as Q3.

Answer. Q4. Sketch the graph of

Answer. On plotting these points, we obtain the graph of $y=|x+3|$ as follows. Q5. Find the area bounded by curve

Answer. The graph of y = sin x can be drawn as ∴ Required area = Area OABO + Area BCDB

Q6. Find the area enclosed between the parabola ${y}^{2}$ = 4ax and the line y = mx

Answer. The area enclosed between the parabola, \(y^{2} = 4ax, and the line, y = mx, is represented by the shaded area OABO as ∴ Area OABO = Area OCABO – Area (ΔOCA) Q7. Find the area enclosed by the parabola

Answer. The area enclosed between the parabola  is represented by the shaded area OBAO as The points of intersection of the given curves are A (–2, 3) and (4, 12). We draw AC and BD perpendicular to x-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO) $\begin{array}{rl}& ={\int }_{-2}^{4}\frac{1}{2}\left(3x+12\right)dx-{\int }_{-2}^{1}\frac{3{x}^{2}}{4}dx\\ & =\frac{1}{2}{\left[\frac{3{x}^{2}}{2}+12x\right]}_{-2}^{4}-\frac{3}{4}{\left[\frac{{x}^{3}}{3}\right]}_{-2}^{4}\end{array}$

Q8.

Answer. The area of the smaller region bounded by the ellipse,  is represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO) $\begin{array}{rl}& ={\int }_{0}^{3}2\sqrt{1-\frac{{x}^{2}}{9}}dx-{\int }_{0}^{3}2\left(1-\frac{x}{3}\right)dx\\ & =\frac{2}{3}\left[{\int }_{0}^{3}\sqrt{9-{x}^{2}}dx\right]-\frac{2}{3}{\int }_{0}^{3}\left(3-x\right)dx\\ & =\frac{2}{3}{\left[\frac{x}{2}\sqrt{9-{x}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{x}{3}\right]}_{0}^{3}-\frac{2}{3}{\left[3x-\frac{{x}^{2}}{2}\right]}_{0}^{3}\end{array}$

Q9. Find the area of the smaller region bounded by the Find the area of the smaller region bounded by the ellipse

Answer. The area of the smaller region bounded by the ellipse,  is represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO) $\begin{array}{rl}& ={\int }_{0}^{a}b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}dx-{\int }_{b}^{a}b\left(1-\frac{x}{a}\right)dx\\ & =\frac{b}{a}{\int }_{0}^{0}\sqrt{{a}^{2}-{x}^{2}}dx-\frac{b}{a}{\int }_{0}^{a}\left(a-x\right)dx\end{array}$

Q10. Find the area of the region enclosed by the parabola ${x}^{2}$ = y, the line y = x + 2 and x-axis

Answer. The area of the region enclosed by the parabola, ${x}^{2}$ = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as Q11.