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NCERT Solutions Class 12 Maths Chapter 7 - Integrals

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 Page 1


1 	 / 	 1 5
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 	 - 	 E x e r c i s e 	 7 9 .
E v a l u a t e 	 t h e 	 i n t e g r a l s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 8 	 u s i n g 	 s u b s t i t u t i o n s .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	
= 	 	 … … … . ( i )
P u t t i n g 	
	 	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 x 	 t o 	 t 	
w h e n 	 x 	 = 	 0 , 	 t 	 = 	 x
2
	 + 1 	 = 	 0 	 + 1 	 = 	 1
w h e n 	 x 	 = 1 , 	 t 	 = 	 x
2
	 + 1 	 = 	 1 	 + 1 	 = 	 2
	 	 F r o m 	 e q . 	 ( i ) , 	
	
= 	
Page 2


1 	 / 	 1 5
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 	 - 	 E x e r c i s e 	 7 9 .
E v a l u a t e 	 t h e 	 i n t e g r a l s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 8 	 u s i n g 	 s u b s t i t u t i o n s .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	
= 	 	 … … … . ( i )
P u t t i n g 	
	 	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 x 	 t o 	 t 	
w h e n 	 x 	 = 	 0 , 	 t 	 = 	 x
2
	 + 1 	 = 	 0 	 + 1 	 = 	 1
w h e n 	 x 	 = 1 , 	 t 	 = 	 x
2
	 + 1 	 = 	 1 	 + 1 	 = 	 2
	 	 F r o m 	 e q . 	 ( i ) , 	
	
= 	
2 	 / 	 1 5
= 	
= 	 	
= 	
= 	 	 A n s . 	
2 . 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 	 t o 	 	
W h e n 	 	
W h e n 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
Page 3


1 	 / 	 1 5
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 	 - 	 E x e r c i s e 	 7 9 .
E v a l u a t e 	 t h e 	 i n t e g r a l s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 8 	 u s i n g 	 s u b s t i t u t i o n s .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	
= 	 	 … … … . ( i )
P u t t i n g 	
	 	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 x 	 t o 	 t 	
w h e n 	 x 	 = 	 0 , 	 t 	 = 	 x
2
	 + 1 	 = 	 0 	 + 1 	 = 	 1
w h e n 	 x 	 = 1 , 	 t 	 = 	 x
2
	 + 1 	 = 	 1 	 + 1 	 = 	 2
	 	 F r o m 	 e q . 	 ( i ) , 	
	
= 	
2 	 / 	 1 5
= 	
= 	 	
= 	
= 	 	 A n s . 	
2 . 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 	 t o 	 	
W h e n 	 	
W h e n 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
3 	 / 	 1 5
I 	 = 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
Page 4


1 	 / 	 1 5
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 	 - 	 E x e r c i s e 	 7 9 .
E v a l u a t e 	 t h e 	 i n t e g r a l s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 8 	 u s i n g 	 s u b s t i t u t i o n s .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	
= 	 	 … … … . ( i )
P u t t i n g 	
	 	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 x 	 t o 	 t 	
w h e n 	 x 	 = 	 0 , 	 t 	 = 	 x
2
	 + 1 	 = 	 0 	 + 1 	 = 	 1
w h e n 	 x 	 = 1 , 	 t 	 = 	 x
2
	 + 1 	 = 	 1 	 + 1 	 = 	 2
	 	 F r o m 	 e q . 	 ( i ) , 	
	
= 	
2 	 / 	 1 5
= 	
= 	 	
= 	
= 	 	 A n s . 	
2 . 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 	 t o 	 	
W h e n 	 	
W h e n 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
3 	 / 	 1 5
I 	 = 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
4 	 / 	 1 5
= 	
= 	 	 A n s . 	
3 . 	 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
L i m i t s 	 o f 	 i n t e g r a t i o n , 	 w h e n 	 	 	
w h e n 	
	
	 	 F r o m 	 e q . 	 ( i ) ,
I 	 = 	
= 	
Page 5


1 	 / 	 1 5
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 	 - 	 E x e r c i s e 	 7 9 .
E v a l u a t e 	 t h e 	 i n t e g r a l s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 8 	 u s i n g 	 s u b s t i t u t i o n s .
1 . 	 	
A n s . 	 L e t 	 I 	 = 	
= 	 	 … … … . ( i )
P u t t i n g 	
	 	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 x 	 t o 	 t 	
w h e n 	 x 	 = 	 0 , 	 t 	 = 	 x
2
	 + 1 	 = 	 0 	 + 1 	 = 	 1
w h e n 	 x 	 = 1 , 	 t 	 = 	 x
2
	 + 1 	 = 	 1 	 + 1 	 = 	 2
	 	 F r o m 	 e q . 	 ( i ) , 	
	
= 	
2 	 / 	 1 5
= 	
= 	 	
= 	
= 	 	 A n s . 	
2 . 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
T o 	 c h a n g e 	 t h e 	 l i m i t s 	 o f 	 i n t e g r a t i o n 	 f r o m 	 	 t o 	 	
W h e n 	 	
W h e n 	 	
	 	 F r o m 	 e q . 	 ( i ) ,
3 	 / 	 1 5
I 	 = 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
= 	
4 	 / 	 1 5
= 	
= 	 	 A n s . 	
3 . 	 	
A n s . 	 L e t 	 I 	 = 	 	 … … … . ( i )
P u t t i n g 	
	
	 	
L i m i t s 	 o f 	 i n t e g r a t i o n , 	 w h e n 	 	 	
w h e n 	
	
	 	 F r o m 	 e q . 	 ( i ) ,
I 	 = 	
= 	
5 	 / 	 1 5
= 	
= 	 	
[ A p p l y i n g 	 P r o d u c t 	 R u l e ]
= 	
= 	
= 	
= 	 	
= 	
= 	
= 	
= 	 	 A n s . 	
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FAQs on NCERT Solutions Class 12 Maths Chapter 7 - Integrals

1. What are integrals?
Ans. Integrals are mathematical tools used to calculate the area under a curve. They are also used to find the accumulation of quantities and solve various mathematical problems.
2. How are integrals used in real-life applications?
Ans. Integrals have numerous real-life applications, such as calculating the total distance traveled by an object, finding the average value of a function, determining the area of irregular shapes, and analyzing data in fields like physics, economics, and engineering.
3. What is the difference between definite and indefinite integrals?
Ans. Definite integrals have upper and lower limits and give a specific value as the result. They are used to find the accumulated amount of a quantity within a given interval. On the other hand, indefinite integrals do not have limits and yield a function as the result. They are used to find antiderivatives.
4. How can we solve integrals?
Ans. Integrals can be solved using various techniques, such as integration by substitution, integration by parts, trigonometric substitution, and partial fractions. The choice of technique depends on the complexity of the integrand and the available knowledge of integration rules.
5. Can integrals be used to solve differential equations?
Ans. Yes, integrals can be used to solve differential equations. Differential equations involve derivatives, and integrating both sides of the equation can help find their solutions. This process often leads to finding an antiderivative and applying initial conditions to determine the specific solution.
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