Mathematics Multiple Choice Questions and Answers for Class 12 on “Fundamental Theorem of Calculus-2”.
1. Evaluate the integral (int_1^5x^2 ,dx).
a) (frac{125}{3})
b) (frac{124}{3})
c) 124
d) –(frac{124}{3})
Answer: b
Clarification: Let I=(int_1^5x^2 ,dx)
F(x)=(int x^2 ,dx)
=(frac{x^3}{3})
By using the fundamental theorem of calculus, we get
I=F(5)-F(1)
=(frac{5^3}{3}-frac{1^3}{3}=frac{125-1}{3}=frac{124}{3})
2. Find (int_{π/4}^{π/2}7 ,cosx ,dx).
a) 7(1-(frac{1}{sqrt{2}}))
b) -7(1-(frac{1}{sqrt{2}}))
c) 7(1+(frac{1}{sqrt{2}}))
d) 7((sqrt{2}-frac{1}{sqrt{2}}))
Answer: a
Clarification: Let (I=int_{π/4}^{π/2}7 ,cosx ,dx)
F(x)=∫ 7 cosx dx
=7(sinx)
Applying the limits by using the second fundamental theorem of calculus, we get
(I=F(frac{π}{2})-F(frac{π}{4})=7(sinfrac{π}{2}-sin frac{π}{4})=7(1-frac{1}{sqrt{2}}))
3. The value of the integral (int_0^1(x+3) ,e^{3x} ,dx).
a) (frac{8e^3}{9})
b) (frac{11}{9} e^3-8)
c) (frac{e^{3x}}{9}(x+8))
d) (frac{11}{9} e^3-frac{8}{9})
Answer: d
Clarification: Let (I=int_0^1(x+3) ,e^{3x} ,dx)
F(x)=(int (x+3) ,e^{3x} ,dx)
By using the formula (int ,u.v ,dx=uint ,v dx-int ,u'(int ,v ,dx)), we get
F(x)=(x+3) (int e^{3x} ,dx-int ,(x+3)’int ,e^{3x} ,dx)
=(frac{(x+3) ,e^{3x}}{3}-int frac{e^{3x}}{3} dx)
=(frac{(x+3) e^{3x}}{3}-frac{e^{3x}}{9})
=(frac{e^{3x}}{3} (x+3-frac{1}{3})=frac{e^{3x}}{9}(3x+8))
Applying the limits, we get
I=F(1)-F(0)
=(frac{e^{3(1)}}{9} (3+8)-frac{e^{3(0)}}{9}(0+8))
=(frac{11}{9} e^3-frac{8}{9}).
4. Find (int_0^π(1-sin3x)dx).
a) (frac{3π-2}{4})
b) 3π-1
c) (frac{3π-2}{3})
d) π-(frac{1}{3})
Answer: c
Clarification: Let (I=int_0^π(1-sin3x)dx)
F(x)=∫ 1-sin3x dx
=x+(frac{cos3x}{3})
Applying the limits by using the fundamental theorem of calculus, we get
I=F(π)-F(0)
=(π+frac{cos3π}{3}-0-frac{cos0}{3})
=(π-frac{1}{3}-frac{1}{3}=π-frac{2}{3}=frac{3π-2}{3}).
5. Evaluate the integral (int_1^{sqrt{3}} frac{3}{1+x^2}).
a) (frac{π}{2})
b) (frac{π}{4})
c) (frac{π}{3})
d) (frac{π}{6})
Answer: b
Clarification: Let (I=int_1^{√3} frac{3}{1+x^2})
F(x)=(int frac{3}{1+x^2}dx)
=3(int frac{1}{1+x^2} ,dx)
=3 tan-1x
Applying the limits, we get
I=F((sqrt{3}))-F(1)
=3 tan-1(sqrt{3})-3 tan-11
(3(frac{π}{3})-frac{3π}{4}=frac{4π-3π}{4}=frac{π}{4}).
6. Find (int_3^45x^3 ,dx).
a) –(frac{185}{4})
b) –(frac{185}{3})
c) (frac{185}{2})
d) (frac{185}{4})
Answer: d
Clarification: Let (I=int_3^45x^3 ,dx)
F(x)=∫ 5x3 dx
=(frac{5x^4}{4})
Applying the limits by using the fundamental theorem of calculus, we get
I=F(4)-F(3)
=(frac{5(4)^3}{4}-frac{5(3)^3}{4}=frac{5}{4}(4^3-3^3))
=(frac{5}{4} (64-27)=frac{5}{4} (37)=frac{185}{4})
7. Evaluate the definite integral (int_0^1 sin^2x ,dx).
a) –(frac{π}{2})
b) π
c) (frac{π}{4})
d) (frac{π}{6})
Answer: c
Clarification: Let (I=int_0^{π/2}sin^{2}x ,dx)
F(x)=(int sin^2x ,dx)
=(int frac{(1-cos2x)}{2} ,dx)
=(frac{1}{2} (x-frac{sin2x}{2}))
Applying the limits, we get
(I=F(frac{π}{2})-F(0)=frac{1}{2} (frac{π}{2}-frac{sinπ}{2})-frac{1}{2} (0-frac{sin0}{2}))
=(frac{1}{4} (π-0)-0=frac{π}{4}).
8. Find (int_1^2sqrt{x}-3x ,dx).
a) (frac{8sqrt{2}-31}{6})
b) (8sqrt{2}-31)
c) (frac{sqrt{2}-31}{3})
d) (frac{8sqrt{2}+31}{4})
Answer: a
Clarification: Let (I=int_1^2 sqrt{x}-3x ,dx)
F(x)=(int sqrt{x}-3x ,dx)
=(frac{x^{1/2+1}}{1/2+1}-frac{3x^2}{2}=frac{2x^{frac{3}{2}}}{3}-frac{3x^2}{2})
By using the second fundamental theorem of calculus, we get
I=F(2)-F(1)=(left(frac{2×2^{3/2}}{3}-frac{3×2^2}{2}right)-left(frac{2×1^{3/2}}{3}-frac{3×1^2}{2}right))
I=(frac{4sqrt{2}}{3}-6-frac{2}{3}+frac{3}{2}=frac{8sqrt{2}-36-4+9}{6}=frac{8sqrt{2}-31}{6})
9. Find the value (int_{-1}^23x+x^2-2 ,dx).
a) –(frac{4}{3})
b) (frac{3}{2})
c) (frac{5}{6})
d) –(frac{5}{6})
Answer: b
Clarification: Let (I=int_{-1}^23x+x^2-2 ,dx)
F(x)=(int 3x+x^2-2 ,dx)
=(frac{3x^2}{2}+frac{x^3}{3}-2x)
Applying the limits, we get
I=F(2)-F(-1)
I=(left(frac{(3×2^3)}{2}+frac{2^3}{3}-2(2)right)-left(frac{3 (-1)^2}{2}+frac{(-1)^3}{3}-2(-1)right))
I=(6+frac{8}{3}-4-frac{3}{2}+frac{1}{3}-2=frac{3}{2}).
10. Find (int_1^2 logx.x^2 dx)
a) log2-(frac{7}{3})
b) (frac{8}{3}) log2-5
c) (frac{8}{3}) log2-log3
d) (frac{8}{3}) log2
Answer: b
Clarification: (I=int_0^1 logx.x^2 dx)
F(x)=(int logx.x^2 dx)
By using the formula (int u.v dx=uint v dx-int u'(int v dx)), we get
(int logx.x^2 ,dx=logx int x^2 dx-int (logx)’ int ,x^2 dx)
=(frac{x^3 logx}{3}-int frac{1}{x}.x^3/3 dx)
∴(F(x)=frac{x^3 logx}{3}-frac{x^3}{9}=frac{x^3}{3} (logx-frac{1}{3}))
Hence, by using the fundamental theorem of calculus, we get
I=F(2)-F(1)
I=(frac{2^3}{3} ,(log2-frac{2}{3})-frac{1^3}{3} ,(log1-frac{1}{3}))
I=(frac{2^3}{3} ,log2-frac{16}{3}+frac{1}{3})
I=(frac{8}{3}) log2-5
Mathematics Multiple Choice Questions and Answers for Class 12,