Mathematics Multiple Choice Questions on “Fundamental Theorem of Calculus-1”.
1. Find (int_0^8x ,dx).
a) 32
b) 34
c) 21
d) 24
Answer: a
Clarification: Let I=(int_0^8x ,dx)
F(x)=(int x ,dx=frac{x^2}{2})
Using the second fundamental theorem of calculus, we get
I=F(8)-f(0)
∴(int_0^8x ,dx=frac{8^2}{2}-0=32)
2. Find (int_0^{frac{π}{2}} ,5 ,sinx ,dx).
a) -5
b) 9
c) 5
d) -9
Answer: c
Clarification: Let (I=int_0^{frac{π}{2}} ,5 ,sinx ,dx)
F(x)=(int5 ,sinx ,dx=-5 ,cosx)
Applying the limits by using the fundamental theorem of calculus, we get
I=F((frac{π}{2}))-F(0)
∴(int_0^{frac{π}{2}} ,5 ,sinx ,dx=-5[cosfrac{π}{2}-cos0])
=-5[0-1]=5
3. Find the value of (int_4^5 ,logx ,dx).
a) 5 log5-log4+1
b) 5 log5-4 log4-1
c) 4 log5-4 log4-1
d) 5-4 log4-log5
Answer: b
Clarification: Let I=(int_4^5 ,logx ,dx).
F(x)=∫ logx dx
By using the formula (int ,u.v dx=u int v ,dx-int u'(int ,v ,dx)), we get
(int log x ,dx=logx int ,dx-int(logx)’int ,dx)
F(x)=x logx-∫ dx=x(logx-1).
Applying the limits using the fundamental theorem of calculus, we get
I=F(5)-F(4)=(5 log5-5)-(4 log4-4)
=5 log5-4 log4-1.
4. Find (int_0^{frac{π}{4}} ,9 ,cos^2x ,dx).
a) (frac{9}{2}left (frac{π}{6}-1right))
b) (frac{9}{4}left (frac{π}{2}+1right))
c) (frac{9}{4}left (frac{π}{2}-1right))
d) (left (frac{π}{2}-1right))
Answer: c
Clarification: Let I=(int_0^{frac{π}{4}} ,9 ,cos^2x ,dx).
F(x)=(int ,9 ,cos^2x ,dx)
=9(int(frac{1+cos2x}{2})dx)
=(frac{9}{2} (x-frac{sin2x}{2}))
Applying the limits, we get
I=(F(frac{π}{4})-F(0)=frac{9}{2} left (frac{π}{4}-frac{sin2(frac{π}{4})}{2}right)-frac{9}{2} (0-frac{sin0}{2}))
=(frac{9}{2}left (frac{π}{4}-frac{sinπ/2}{2}right )=frac{9}{4} (π/2-1))
5. Find (int_0^2 ,e^{2x} ,dx).
a) (frac{e^4-1}{6})
b) (frac{e^4+1}{2})
c) (frac{e-1}{2})
d) (frac{e^4-1}{2})
Answer: d
Clarification: Let (I=int_0^2 ,e^2x ,dx)
F(x)=(int e^{2x} dx)
=(frac{e^{2x}}{2})
Applying the limits, we get
I=F(2)-F(0)
=(frac{e^2(2)}{2}-frac{e^2(0)}{2}=frac{(e^4-1)}{2}).
6. Find (int_{π/4}^{π/2} ,2sinx ,sin(cosx) ,dx).
a) 2(1-cos(frac{1}{sqrt{2}}))
b) (cos(frac{1}{sqrt{2}})-cos1)
c) 2(cos(frac{1}{sqrt{2}})+1)
d) (cos(frac{1}{sqrt{2}})+cos1)
Answer: a
Clarification: Let (I=int_{π/4}^{π/2} ,2sinx ,sin(cosx) ,dx)
F(x)=(int 2 ,sinx ,sin(cosx)dx)
Let cosx=t
Differentiating w.r.t x, we get
sinx dx=dt
∴(int 2 ,sinx ,sin(cosx)dx=int 2 ,sint ,dt=-2 ,cost)
Replacing t with cosx, we get
∴∫ 2 sinx sin(cosx)dx=-2 cos(cosx)
By applying the limits, we get
(I=F(frac{π}{4})-F(frac{π}{2})=-2 cos(frac{cosπ}{4})+2 cos(frac{cosπ}{2}))
=2(1-cos(frac{1}{sqrt{2}}))
7. Find (int_{-2}^1 ,5x^4 ,dx).
a) 54
b) 75
c) 33
d) 36
Answer: c
Clarification: (I=int_{-2}^1 ,5x^4 ,dx)
F(x)=(int5x^4 ,dx=5(frac{x^5}{5})=x^5)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(1)-F(-2)
=(1)5-(-2)5=1+32=33.
8. Find (int_0^3 ,e^x ,dx).
a) e3+1
b) -e3-1
c) e3-1
d) 3e3-2
Answer: c
Clarification: Let I=(int_0^3 ,e^x ,dx)
F(x)=(int ,e^x ,dx=e^x)
Applying the limits, we get
I=F(3)-F(0)
=e3-e0=e3-1.
9. Find (int_0^{π/4} ,2 ,tanx ,dx).
a) log2
b) log(sqrt{2})
c) 2 log2
d) 0
Answer: a
Clarification: (I=int_0^{π/4} ,2 ,tanx ,dx)
F(x)=∫ 2 tanx dx
=2∫ tanx dx
=2 log|secx|
Therefore, by using the fundamental theorem of calculus, we get
I=F(π/4)-F(0)
(=2left(log|sec frac{π}{4}|-log|sec0|right)=2 logsqrt{2}-log1)
(=2 logsqrt{2}=log(sqrt{2})^2=log2)
I=(frac{8}{3} log2-frac{8}{3}-0+frac{1}{3}=frac{8}{3} log2-frac{7}{3}).
10. Find (int_{-1}^1 ,2xe^x ,dx).
a) (frac{4}{e})
b) 4e
c) –(frac{4}{e})
d) -4e
Answer: a
Clarification: (I=int_{-1}^1 ,2xe^x ,dx)
F(x)=(int 2xe^x dx)
By using the formula, (int u.v ,dx=u int v ,dx-int u'(int v ,dx))
F(x)=2x(int e^x dx-int(2x)’int e^x ,dx)
=(2xe^x-int 2e^x dx)
=(2e^x (x-1))
Therefore, by using the fundamental theorem of calculus, we get
I=F(1)-F(-1)
I=2e1 (1-1)-2e-1 (-1-1)
I=(frac{4}{e}).