12th Grade > Mathematics
DEFINITE INTEGRATION MCQs
Total Questions : 30
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Answer: Option C. -> 812
:
C
The given equation of curve can be written as x=f(y)=9y−y3.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(9−y2)
f(y)=y.(3+y)(3−y)
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=∫0−3|f(y)|dy+∫30f(y)dyA=∫0−3|9y−y3|dy+∫30(9y−y3)dyA=−∫0−3(9y−y3)dy+∫309y−y3dyA=−[9y22−y44]0−3+[9y22−y44]30A=−[0−0−812+814]+[812−814]A=81−812A=812
:
C
The given equation of curve can be written as x=f(y)=9y−y3.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(9−y2)
f(y)=y.(3+y)(3−y)
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=∫0−3|f(y)|dy+∫30f(y)dyA=∫0−3|9y−y3|dy+∫30(9y−y3)dyA=−∫0−3(9y−y3)dy+∫309y−y3dyA=−[9y22−y44]0−3+[9y22−y44]30A=−[0−0−812+814]+[812−814]A=81−812A=812
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(√4ax−mx)dx
∴∫4am20(√4ax−mx)dx=a23
⇒83a2m3=a23⇒m3=8⇒m=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(√4ax−mx)dx
∴∫4am20(√4ax−mx)dx=a23
⇒83a2m3=a23⇒m3=8⇒m=2
Answer: Option B. -> e - 1
:
B
limπ→∞∑nr=11nern=∫10exdx=[ex]10=e−1
:
B
limπ→∞∑nr=11nern=∫10exdx=[ex]10=e−1
Answer: Option B. -> π log 2
:
B
LetI=∫∞0log(1+x2)1+x2dx
Putx=tanθ⇒dx=sec2θdθ,
∴I=∫n20log(secθ)2dθ=2∫n20logsecθdθ
=−2∫n20logcosθdθ =−2.π2log12=−πlog12=πlog2
:
B
LetI=∫∞0log(1+x2)1+x2dx
Putx=tanθ⇒dx=sec2θdθ,
∴I=∫n20log(secθ)2dθ=2∫n20logsecθdθ
=−2∫n20logcosθdθ =−2.π2log12=−πlog12=πlog2
Answer: Option D. -> −√2−√3+5
:
D
∫20[x2]dx
=∫10[x2]dx+∫√20[x2]dx+∫√3√2[x2]dx+∫2√3[x2]dx
=∫100dx+∫√201dx+∫√3√22dx+∫2√33dx
=√2−1+2√3−2√2+6−3√3
=5−√3−√2
:
D
∫20[x2]dx
=∫10[x2]dx+∫√20[x2]dx+∫√3√2[x2]dx+∫2√3[x2]dx
=∫100dx+∫√201dx+∫√3√22dx+∫2√33dx
=√2−1+2√3−2√2+6−3√3
=5−√3−√2
Answer: Option B. -> 215
:
B
Using gamma function,
∫π20sin2xcos3xdx=r(32)r22r(72)=215
:
B
Using gamma function,
∫π20sin2xcos3xdx=r(32)r22r(72)=215
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.
:
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.
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 -
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).
∫3−3f(x)dx=∫1−3f(x)dx+∫31f(x)dx
And given that for (−3,1)f(x)=x2
And for (1,3) f(x) = 6-x
∫3−3f(x)dx=∫1−3(x2)dx+∫31(6−x)dx
This way we can calculate the area even if the function is discontinuous.
:
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 -
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).
∫3−3f(x)dx=∫1−3f(x)dx+∫31f(x)dx
And given that for (−3,1)f(x)=x2
And for (1,3) f(x) = 6-x
∫3−3f(x)dx=∫1−3(x2)dx+∫31(6−x)dx
This way we can calculate the area even if the function is discontinuous.
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=lima→∞∫a11x2dx
= lim a→∞(−1x)|a1 Since, ∫1x2dx=(−1x)
=0−(−1)=1
:
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=lima→∞∫a11x2dx
= lim a→∞(−1x)|a1 Since, ∫1x2dx=(−1x)
=0−(−1)=1
Answer: Option C. -> 12
:
C
I1=∫k1−kxf{x(1−x)}dx
=∫k1−k(1−k+k−x)f[(1−k+k−x){1−(1−k+k−x)}]dx
=∫k1−k(1−x)f{x(1−x)}dx
=∫k1−kf{x(1−x)}dx−∫k1−kxf{x(1−x)}dx=I2−I1
∴2I1=I2⇒I1I2=12
:
C
I1=∫k1−kxf{x(1−x)}dx
=∫k1−k(1−k+k−x)f[(1−k+k−x){1−(1−k+k−x)}]dx
=∫k1−k(1−x)f{x(1−x)}dx
=∫k1−kf{x(1−x)}dx−∫k1−kxf{x(1−x)}dx=I2−I1
∴2I1=I2⇒I1I2=12