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

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.
Solutions Class 12 Maths (Application Of Integrals) Miscellaneous Exercise

Q1. 

Answer. (i) The required area is represented by the shaded area ADCBA as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  Area ADCBA=2ydx=x2dx=[x33]12=8313=73 units  (ii) The required area is represented by the shaded area ADCBA as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise 

Q2. Find the area between the curve 

Answer. The required area is represented by the shaded area OBAO as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  The points of intersection of the curves, y=x and y=x2, is A(1,1) .  We draw AC perpendicular to x -axis.   Area (OBAO)= Area (ΔOCA) Area (OCABO) (i) 

Q3. 

Answer.  The area in the first quadrant bounded by y=4x2,x=0,y=1, and y=4 is  represented by the shaded area ABCDA as  Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  Area ABCD=x1xdx=y2dx 

Q4. Sketch the graph of y=|x+3| and evaluate 60|x+3|dx

Answer.  The given equation is y=|x+3| The corresponding values of x and y are given in the following table.  Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise On plotting these points, we obtain the graph of y=|x+3| as follows. Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  It is known that, (x+3)0 for 6x3 and (x+3)0 for 3x060(x+3)dx=63(x+3)dx+30(x+3)dx=[x22+3x]63+[x22+3x]30 

Q5. Find the area bounded by curve 

Answer. The graph of y = sin x can be drawn as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise ∴ Required area = Area OABO + Area BCDB 

Q6. Find the area enclosed between the parabola y2 = 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 Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  The points of intersection of both the curves are (0,0) and (4dm2,4am) we draw AC perpendicular to x -axis.  ∴ Area OABO = Area OCABO – Area (ΔOCA) Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise 

Q7. Find the area enclosed by the parabola 4y=3x2 and the line 2y=3x+12

Answer. The area enclosed between the parabola 4y=3x2 and the line 2y=3x+12 is represented by the shaded area OBAO as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise 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) =2412(3x+12)dx213x24dx=12[3x22+12x]2434[x33]24 

Q8. 

Answer. The area of the smaller region bounded by the ellipse, x29+y24=1 and the line, x3+y2=1 is represented by the shaded region BCAB as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise ∴ Area BCAB = Area (OBCAO) – Area (OBAO) =0321x29dx032(1x3)dx=23[039x2dx]2303(3x)dx=23[x29x2+92sin1x3]0323[3xx22]03 

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, x2a2+y2b2=1 and the line xa+yb=1 is represented by the shaded region BCAB as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise ∴ Area BCAB = Area (OBCAO) – Area (OBAO) =0ab1x2a2dxbab(1xa)dx=ba00a2x2dxba0a(ax)dx 

Q10. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis

Answer. The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as Solutions Class 12 maths Chapter-8 (Application Of Integrals) Miscellaneous Exercise  The point of intersection of the parabola, x2=y, and the line, y=x+2, is A(1,1) .  Area OABCO = Area (BCA)+ Area COAC  

Q11.  Using the method of integration find the area bounded by the curve |x|+|y|=1 [Hint: the required region is bounded by lines x+y=1,xy=1,x+y=1 and xy=11]

Answer.