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NCERT Solutions Class 12 Maths Chapter 7 - Integrals

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 Page 1


1 	 / 	 5 0
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 M i s c e l l a n e o u s 	 E x e r c i s e
I n t e g r a t e 	 t h e 	 f u n c t i o n 	 i n 	 E x e r c i s e s 	 1 	 t o 	 1 1 .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	 	
H e r e , 	 	
= 	
= 	
= 	 … … … . ( i )
	 	 	
	 	
	 	
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 – A 	 + 	 B 	 – 	 C 	 = 	 0 	 … … … . ( i i )
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 B 	 + 	 C 	 = 	 0 	 … … … . ( i i i )
C o m p a r i n g 	 c o n s t a n t s 	 A 	 = 	 1 	 … … … . ( i v )
Page 2


1 	 / 	 5 0
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 M i s c e l l a n e o u s 	 E x e r c i s e
I n t e g r a t e 	 t h e 	 f u n c t i o n 	 i n 	 E x e r c i s e s 	 1 	 t o 	 1 1 .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	 	
H e r e , 	 	
= 	
= 	
= 	 … … … . ( i )
	 	 	
	 	
	 	
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 – A 	 + 	 B 	 – 	 C 	 = 	 0 	 … … … . ( i i )
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 B 	 + 	 C 	 = 	 0 	 … … … . ( i i i )
C o m p a r i n g 	 c o n s t a n t s 	 A 	 = 	 1 	 … … … . ( i v )
2 	 / 	 5 0
O n 	 s o l v i n g 	 e q . 	 ( i i ) , 	 ( i i i ) 	 a n d 	 ( i v ) , 	 w e 	 g e t 	 	 A 	 = 	 1 , 	 B 	 = 	 	 C 	 = 	 	
P u t t i n g 	 t h e s e 	 v a l u e s 	 i n 	 e q . 	 ( i ) ,
= 	
	
= 	
= 	 	
= 	
= 	
= 	
= 	 	 A n s .
2 . 	
A n s . 	 L e t 	 I 	 = 	
Page 3


1 	 / 	 5 0
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 M i s c e l l a n e o u s 	 E x e r c i s e
I n t e g r a t e 	 t h e 	 f u n c t i o n 	 i n 	 E x e r c i s e s 	 1 	 t o 	 1 1 .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	 	
H e r e , 	 	
= 	
= 	
= 	 … … … . ( i )
	 	 	
	 	
	 	
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 – A 	 + 	 B 	 – 	 C 	 = 	 0 	 … … … . ( i i )
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 B 	 + 	 C 	 = 	 0 	 … … … . ( i i i )
C o m p a r i n g 	 c o n s t a n t s 	 A 	 = 	 1 	 … … … . ( i v )
2 	 / 	 5 0
O n 	 s o l v i n g 	 e q . 	 ( i i ) , 	 ( i i i ) 	 a n d 	 ( i v ) , 	 w e 	 g e t 	 	 A 	 = 	 1 , 	 B 	 = 	 	 C 	 = 	 	
P u t t i n g 	 t h e s e 	 v a l u e s 	 i n 	 e q . 	 ( i ) ,
= 	
	
= 	
= 	 	
= 	
= 	
= 	
= 	 	 A n s .
2 . 	
A n s . 	 L e t 	 I 	 = 	
3 	 / 	 5 0
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	 	 A n s . 	
3 . 	 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
Page 4


1 	 / 	 5 0
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 M i s c e l l a n e o u s 	 E x e r c i s e
I n t e g r a t e 	 t h e 	 f u n c t i o n 	 i n 	 E x e r c i s e s 	 1 	 t o 	 1 1 .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	 	
H e r e , 	 	
= 	
= 	
= 	 … … … . ( i )
	 	 	
	 	
	 	
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 – A 	 + 	 B 	 – 	 C 	 = 	 0 	 … … … . ( i i )
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 B 	 + 	 C 	 = 	 0 	 … … … . ( i i i )
C o m p a r i n g 	 c o n s t a n t s 	 A 	 = 	 1 	 … … … . ( i v )
2 	 / 	 5 0
O n 	 s o l v i n g 	 e q . 	 ( i i ) , 	 ( i i i ) 	 a n d 	 ( i v ) , 	 w e 	 g e t 	 	 A 	 = 	 1 , 	 B 	 = 	 	 C 	 = 	 	
P u t t i n g 	 t h e s e 	 v a l u e s 	 i n 	 e q . 	 ( i ) ,
= 	
	
= 	
= 	 	
= 	
= 	
= 	
= 	 	 A n s .
2 . 	
A n s . 	 L e t 	 I 	 = 	
3 	 / 	 5 0
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	 	 A n s . 	
3 . 	 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
4 	 / 	 5 0
	 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
I 	 = 	
= 	
= 	 	
= 	
= 	
= 	 	 A n s . 	
4 . 	
A n s . 	 L e t 	 I 	 = 	
Page 5


1 	 / 	 5 0
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 M i s c e l l a n e o u s 	 E x e r c i s e
I n t e g r a t e 	 t h e 	 f u n c t i o n 	 i n 	 E x e r c i s e s 	 1 	 t o 	 1 1 .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	 	
H e r e , 	 	
= 	
= 	
= 	 … … … . ( i )
	 	 	
	 	
	 	
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 – A 	 + 	 B 	 – 	 C 	 = 	 0 	 … … … . ( i i )
C o m p a r i n g 	 t h e 	 c o e f f i c i e n t s 	 o f 	 	 B 	 + 	 C 	 = 	 0 	 … … … . ( i i i )
C o m p a r i n g 	 c o n s t a n t s 	 A 	 = 	 1 	 … … … . ( i v )
2 	 / 	 5 0
O n 	 s o l v i n g 	 e q . 	 ( i i ) , 	 ( i i i ) 	 a n d 	 ( i v ) , 	 w e 	 g e t 	 	 A 	 = 	 1 , 	 B 	 = 	 	 C 	 = 	 	
P u t t i n g 	 t h e s e 	 v a l u e s 	 i n 	 e q . 	 ( i ) ,
= 	
	
= 	
= 	 	
= 	
= 	
= 	
= 	 	 A n s .
2 . 	
A n s . 	 L e t 	 I 	 = 	
3 	 / 	 5 0
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	 	 A n s . 	
3 . 	 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
4 	 / 	 5 0
	 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
I 	 = 	
= 	
= 	 	
= 	
= 	
= 	 	 A n s . 	
4 . 	
A n s . 	 L e t 	 I 	 = 	
5 	 / 	 5 0
= 	
= 	
= 	 	 	
P u t t i n g 	
	 	
	 	
	 	 I 	 = 	
= 	
= 	 	 A n s . 	
5 . 	
A n s . 	 L e t 	 I 	 = 	 	
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FAQs on NCERT Solutions Class 12 Maths Chapter 7 - Integrals

1. What is the definition of an integral?
Ans. An integral is a mathematical concept used to find the area under a curve or to calculate the accumulation of quantities over a given interval. It is denoted by the symbol ∫ and represents the antiderivative of a function.
2. What are the different types of integrals?
Ans. There are mainly two types of integrals: definite integral and indefinite integral. The definite integral calculates the accumulation of quantities over a specific interval, while the indefinite integral finds the antiderivative of a function without any specified interval.
3. How do you evaluate a definite integral?
Ans. To evaluate a definite integral, you need to follow these steps: 1. Find the antiderivative of the integrand. 2. Substitute the upper and lower limits of the interval into the antiderivative. 3. Subtract the value obtained for the lower limit from the value obtained for the upper limit.
4. How is the definite integral related to area?
Ans. The definite integral is closely related to finding the area under a curve. When you evaluate a definite integral, the result represents the area between the curve and the x-axis within the specified interval. The sign of the integral represents whether the area is positive or negative.
5. What is the fundamental theorem of calculus?
Ans. The fundamental theorem of calculus states that if a function f(x) is continuous on an interval [a, b] and F(x) is an antiderivative of f(x) on that interval, then the definite integral of f(x) from a to b is equal to F(b) - F(a). This theorem connects the concepts of differentiation and integration, allowing us to evaluate definite integrals by finding antiderivatives.
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