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12th Grade > Mathematics

DEFINITE INTEGRATION MCQs

Total Questions : 30 | Page 1 of 3 pages
Question 1. Area enclosed by curve y39y+x=0 and Y - axis is -
  1.    92
  2.    9
  3.    812
  4.    81
 Discuss Question
Answer: Option C. -> 812
:
C
The given equation of curve can be written as x=f(y)=9yy3.Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
f(y)=y(9y2)
f(y)=y.(3+y)(3y)
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
A=03|f(y)|dy+30f(y)dyA=03|9yy3|dy+30(9yy3)dyA=03(9yy3)dy+309yy3dyA=[9y22y44]03+[9y22y44]30A=[00812+814]+[812814]A=81812A=812
Question 2. If area bounded by the curves y2=4ax and y=mx is a23  then the value of m is
  1.    2
  2.    -2
  3.    12
  4.    None of these
 Discuss Question
Answer: Option A. -> 2
:
A
The two curves y2 = 4ax and y = mx intersect at (4am2,4am) and the area enclosed by the two curves is given by 4am20(4axmx)dx
4am20(4axmx)dx=a23
83a2m3=a23m3=8m=2
Question 3. limπnr=1 1nern is [AIEEE 2004]
  1.    e + 1
  2.    e - 1
  3.    1 - e
  4.    e
 Discuss Question
Answer: Option B. -> e - 1
:
B
limπnr=11nern=10exdx=[ex]10=e1
Question 4. 0 log(1+x2)1+x2dx=
  1.    π log 12
  2.    π log 2
  3.    2π log 12
  4.    2π log 2
 Discuss Question
Answer: Option B. -> π log 2
:
B
LetI=0log(1+x2)1+x2dx
Putx=tanθdx=sec2θdθ,
I=n20log(secθ)2dθ=2n20logsecθdθ
=2n20logcosθdθ =2.π2log12=πlog12=πlog2
Question 5. 20[x2]dx is (where [.] is greastest integral function
  1.    2−√2
  2.    2+√2
  3.    √2−1
  4.    −√2−√3+5
 Discuss Question
Answer: Option D. -> −√2−√3+5
:
D
20[x2]dx
=10[x2]dx+20[x2]dx+32[x2]dx+23[x2]dx
=100dx+201dx+322dx+233dx
=21+2322+633
=532
Question 6. π20 sin2x cos3x dx= [RPET 1984, 2003]
  1.    0
  2.    215
  3.    415
  4.    None of these
 Discuss Question
Answer: Option B. -> 215
:
B
Using gamma function,
π20sin2xcos3xdx=r(32)r22r(72)=215
Question 7. Antiderivative of all the continuous functions can be written in terms of  elementary functions
  1.    False
  2.    True
  3.    a3ex
  4.    None of these
 Discuss Question
Answer: Option A. -> False
:
A
We have seen that there are some functions, antiderivatives of which cannot be written in terms of elementary functions. Functions like ex2,ex2x,sin(1x),etc. are some of the examples of such functions.
Question 8. If a function f(x) is discontinuous in the interval (a,b) then baf(x)dx never exists.
  1.    False
  2.    True
  3.    a3ex
  4.    None of these
 Discuss Question
Answer: Option A. -> False
:
A
We have seen that there are functions which have discontinuity but still can be integrated. These functions have finite type discontinuities. Please try to understand one thing here, that definite integral gives you the area so area can be calculated by the curve into pieces as well.
For example -
If A Function F(x) Is Discontinuous In The Interval (a,b) Th...
We can see that the given function is discontinuous at x =1. So to integrate this function from (-3,3) we’ll have to integrate it into pieces. One piece will be (-3,1) and another will be (1, 3).
33f(x)dx=13f(x)dx+31f(x)dx
And given that for (3,1)f(x)=x2
And for (1,3) f(x) = 6-x
33f(x)dx=13(x2)dx+31(6x)dx
This way we can calculate the area even if the function is discontinuous.
Question 9. 11x2dx does not have a finite value
  1.    False
  2.    True
  3.    a3ex
  4.    None of these
 Discuss Question
Answer: Option A. -> False
:
A
Let’s calculate its value.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll change the limit which is not finite to a variable and then put the limits.
11x2dx=limaa11x2dx
= lim a(1x)|a1 Since, 1x2dx=(1x)
=0(1)=1
Question 10. Let f be a positive function. Let
I1=k1k xf{x(1x)}dx,  I2=k1kf{x(1x)}dx
when 2k1>0. Then I1I2 is               [IIT 1997 Cancelled]
  1.    2
  2.    k
  3.    12
  4.    1
 Discuss Question
Answer: Option C. -> 12
:
C
I1=k1kxf{x(1x)}dx
=k1k(1k+kx)f[(1k+kx){1(1k+kx)}]dx
=k1k(1x)f{x(1x)}dx
=k1kf{x(1x)}dxk1kxf{x(1x)}dx=I2I1
2I1=I2I1I2=12

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