12th Grade > Mathematics
DEFINITE INTEGRALS AND AREAS MCQs
Total Questions : 60
| Page 3 of 6 pages
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. -> 1100
:
B
limπ→∞199+299+399+⋯⋯n99n100=limπ→∞∑nr=1(r99n100)
=limπ→∞1n∑nr=1(rn)99=∫10x99dx=[x100100]10=1100
:
B
limπ→∞199+299+399+⋯⋯n99n100=limπ→∞∑nr=1(r99n100)
=limπ→∞1n∑nr=1(rn)99=∫10x99dx=[x100100]10=1100
Answer: Option B. -> π2
:
B
∫π2−π2sin2xdx=2∫π20sin2xdx=2r(32)r(12)2r(2+22)=π2
:
B
∫π2−π2sin2xdx=2∫π20sin2xdx=2r(32)r(12)2r(2+22)=π2
Answer: Option B. -> 21
:
B
Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
∫3−2f(x)dx=∫1−23x2dx+∫316dx
Now we can integrate these integrands and put limits.
∫1−23x2dx+∫316dx=(x3)|1−2+(6x)|31=1−(−8)+18−6
= 21
:
B
Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
∫3−2f(x)dx=∫1−23x2dx+∫316dx
Now we can integrate these integrands and put limits.
∫1−23x2dx+∫316dx=(x3)|1−2+(6x)|31=1−(−8)+18−6
= 21
Answer: Option A. -> [0,117]
:
A
f(x)=xx3+16⇒f′(x)=16−2x3(x3+16)2∴f′(x)=0⇒x=2Alsof′′(2)<0⇒ The function f(x)=xx3+16 is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = 117
Therefore by the property m(b−a)≤∫baf(x)dx≤M(b−a)
(where m and M are the smallest and greatest values of function)
⇒0≤∫10xx3+16dx≤117
:
A
f(x)=xx3+16⇒f′(x)=16−2x3(x3+16)2∴f′(x)=0⇒x=2Alsof′′(2)<0⇒ The function f(x)=xx3+16 is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = 117
Therefore by the property m(b−a)≤∫baf(x)dx≤M(b−a)
(where m and M are the smallest and greatest values of function)
⇒0≤∫10xx3+16dx≤117
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 A. -> 12
:
A
∫x0dt=x+∫1xtf(t)dt⇒∫1xtf(t)dt=x−∫x1tf(t)dt
Differentiating w.r.t x, we get f(x)=1+{0−xf(x)}
⇒f(x)=1−xf(x)⇒(1+x)f(x)=1⇒f(x)=11+x
∴f(1)=11+1=12
:
A
∫x0dt=x+∫1xtf(t)dt⇒∫1xtf(t)dt=x−∫x1tf(t)dt
Differentiating w.r.t x, we get f(x)=1+{0−xf(x)}
⇒f(x)=1−xf(x)⇒(1+x)f(x)=1⇒f(x)=11+x
∴f(1)=11+1=12
Answer: Option B. -> Upper limit
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