Students appreciate clear and concise ML Aggarwal Maths for Class 12 Solutions Chapter 8 Integrals Ex 8.15 that guide them through exercises.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.15

Very Short answer type questions (1 to 9) :

Evaluate the following (1 to 21) integrals :

Question 1.
(i) $$\int_1^3$$ (2x – 1) dx
(ii) $$\int_3^5$$ (2 – x) dx
Solution:
(I) Comparing $$\int_1^3$$ (2x – 1) dx with $$\int_a^b$$ f(x) dx
Here, f(x) = 2x – 1; a = 1; b = 3
∴ nh = b – a = 3 – 1 = 2
∴ f(a) = f(1) = 2 . 1 – 1 = 1
f(a + h) = f(1 + h) = 2(1 + h) – 1 = 1 + 2h
f(a + 2h) = f(1 + 2h) = 2(1 + 2h) – 1 = 1 + 4h
………………………………………………
………………………………………………

(ii) We know that $$\int_a^b$$ f(x) dx = $$\ {Lt}_{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h) \ldots .+f(a+\overline{n-1} h)]$$
On comparing $$\int_3^5$$ (2 – x) dx with $$\int_a^b$$ f(x)
Here a = 3 ; b = 5 ;
f(x) = 2 – x
and nh = b – a = 5 – 3 = 2

Question 1 (old).
(i) $$\int_0^5\left(x+\frac{1}{2}\right)$$ dx (ISC 2009)
(ii) $$\int_a^b$$ x dx (NCERT)
Solution:
(i) Comparing $$\int_0^5\left(x+\frac{1}{2}\right)$$ dx with $$\int_a^b$$ f(x) dx
Here f(x) = x + $$\frac{1}{2}$$ ;
a = 0 ; b = 5
∴ nh = b – a = 5
∴ f(a) = f(0) = 0 + $$\frac{1}{2}$$ = $$\frac{1}{2}$$
f(a + h) = f (0 + h) = f(h) = h + $$\frac{1}{2}$$
f(a + 2h) = f(2h) = 2h + $$\frac{1}{2}$$
…………………………………………
…………………………………………

(ii) We know that

Question 2.
(i) $$\int_0^2$$ (x2 + 3) dx (NCERT Exemplar)
(ii) $$\int_2^3$$ x2 dx (NCERT)
Solution:
(i) Here f(x) = x2 + 3 ;
a = 0, b = 2 and nh = 2 – 0 = 2
Now f(0) = 02 + 3 ;
f(0 + h) = h2 + 3 ;
f(0 + 2h) = 22h2 + 3 ………………….

(ii) On comparing $$\int_2^3$$ x2 dx with $$\int_a^b$$ f(x) dx
Here a = 2, b = 3, nh = 3 – 2 = 1,
f(x) = x2
f(a) = f(2) = 22 ;
f(2 + h) = (2 + h)2 ;
f(2 + 2h) = (2 + 2h)2

Question 3.
(i) $$\int_0^3$$ (2x2 – 5) dx
(ii) $$\int_0^3$$ (2x2 + 3x + 5) dx
Solution:
(i) Comparing $$\int_0^3$$ (2x2 – 5) dx with $$\int_a^b$$ f(x) dx
Here f(x) = 2x2 – 5 ;
a = 0 ;
b = 3 ;
nh = b – a = 3
∴ f(a) = f(0) = 2 × 02 – 5 = – 5
f(a + h) = f(h) = 2h2 – 5
f(a + 2h) = f(2h) = 2 (2h)2 – 5
…………………………………….
…………………………………….

(ii) We know that
$$\int_a^b$$ f(x) dx = $$\ {Lt}_{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h) \ldots .+f(a+\overline{n-1} h)]$$
On comparing $$\int_0^3$$ (2x2 + 3x + 5) with $$\int_a^b$$ f(x) dx
We have a = 0 ; b = 3;
f(x) = 2x2 + 3x + 5
and nh = b – a = 3 – 0 = 3

Question 4.
(i) $$\int_1^2$$ (3x2 – 1) dx
(ii) $$\int_1^3$$ (2x2 + 5x) dx
Solution:
(i) Here f(x) = 3x2 – 1 ;
a = 1 ; b = 2
and nh = b – a = 2 – 1 = 1
Now f(1) = 3 . 12 – 1 = 2 ;
f(1 + h) = 3 (1 + h)2 – 1 = 2 + 6h + 3h2 ;
f(1 + 2h) = 3 (1 + 2h)2 – 1
= 2 + 12h + 12h2
and so on,
$$f(1+\overline{n-1} h)$$ = 3 $$(1+\overline{n-1} h)^2$$ – 1
= 2 + 6 (n – 1) h + 3 (n – 1)2 h2
∴ By definition

(ii) Here f(x) = 2x2 + 5x ;
a = 1 ; b = 3 and nh = 3 – 1 = 2
f(1) = 7 ;
f(1 + h) = 2 (1 + h)2 + 5 (1 + h) = 7 + 9h + 2h2 ;
f(1 + 2h) = 2 (1 + 2h)2 + 5 (1 + 2h) = 7 + 18h + 18h2
………………………………………………………………………………..
 = 7 + 9 (n – 1) h + 2 (n – 1)2 h2
∴ By def. we have

Question 5.
(i) $$\int_a^b$$ x2 dx
(ii) $$\int_1^3$$ (3x2 + 2x + 1) dx
Solution:
(i) Comparing $$\int_a^b$$ x2 dx with $$\int_a^b$$ f(x) dx
Here f(x) = x2 ; nh = b – a
f(a) = a2
f(a + h) = (a + h)2
f(a + 2h) = (a + 2h)2
……………………………..
……………………………..
$$f(a+\overline{n-1} h)=(a+\overline{n-1} h)^2$$

= a2 (b – a) + a (b – a)2 + $$\frac{2}{6}$$ (b – a)3
= $$\frac{1}{3}$$ [3a2 (b – a) + 3a (b – a)2 + (b – a)3]
= $$\frac{1}{3}$$ (b – a) [3a2 + 3ab – 3a2 + b2 + a2 – 2ab]
= $$\frac{1}{3}$$ (b – a) (b2 + ab + a2)
= $$\frac{b^3-a^3}{3}$$

(ii) Comparing $$\int_1^3$$ (3x2 + 2x + 1) dx with $$\int_a^b$$ f(x) dx
f(x) = 3x2 + 2x + 1 ;
a = 1 ; b = 3
and nh = b – a = 3 – 1 = 2
∴ f(a) = f(1) = 3 . 12 + 2 . 1 + 1 = 6
f(a + h) = f(1 + h)
= 3 (1 + h)2 + 2 (1 + h) + 1
= 3h2 + 8h + 6
f(a + 2h) = f(1 + 2h)
= 3 (1 + 2h)2 + 2 (1 + 2h) + 2 (1 + 2h) + 1
= 12h2 + 16h + 6
…………………………………………..
……………………………………………

Question 6.
(i) $$\int_0^2$$ ex dx (NCERT)
(ii) $$\int_{-1}^5$$ ex dx
Solution:
(i) We know that
$$\int_a^b$$ f(x) dx = $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h [f(a) + f(a + h) + f(a + 2h) + …………….. + $$f(a+\overline{n-1} h)$$]
Here a = 0 ;
b = 2;
f(x) = ex ;
nh = b – a = 2 – 0 = 2
Thus, $$\int_0^2$$ ex dx = $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h [f(0) + f(h) + f(2h) + ………… + f(n – 1) h]

(ii) Comparing $$\int_{-1}^5$$ ex dx with $$\int_a^b$$ f(x) dx
Here, f(x) = ex ;
a = – 1 ; b = 5 ;
nh = b – a = 5 + 1 = 6
∴ f(a) = f(- 1) = e-1
f(a + h) = f(- 1 + h)
= e-1 + h
f(a + 2h) = f(- 1 + 2h)
= e-1 + 2h
…………………………..
…………………………..

Question 6 (old).
$$\int_1^4$$ (3x2 + 2x) dx
Solution:
Here f(x) = 3x2 + 2x ;
a = 1, b = 4
and nh = 4 – 1 = 3
Now f(1) = 3 . 12 + 2 . 1 = 5
f(1 + h) = 3 (1 + h)2 + 2 (1 + h)
= 5 + 8h + 3h2 ;
f(1 + 2h) = 3 (1 + 2h)2 + 2 (1 + 2h)
= 5 + 16h + 12h2 ;
………………………………
………………………………
$$f(1+\overline{n-1} h)$$ = 5 + 8 (n – 1) h + 3 (n – 1)2 h2
∴ By definition, we have

= 5 × 3 + 4 × 3 × 3 + $$\frac{1}{2}$$ × 3 (3) (6)
= 15 + 36 + 27 = 78

Question 7.
$$\int_1^3$$ (x2 + 3x + ex) dx
Solution:
Comparing $$\int_1^3$$ (x2 + 3x + ex) dx with $$\int_a^b$$ f(x) dx f(x) dx, we have
f(x) = x2 + 3x + ex; a = 1; b = 3;
nh = b – a = 3 – 1 = 2
f(a) = f(1) = 12 + 3.1 + e1
f(a + h) = f (1+ h)
= (1+ h)2 +3 (1+ h) + e1 + h
f(a + 2h) = f(1 + 2h)
= (1 + 2h)2 + 3(1 + 2h) + e1 + 2h
…………………………
…………………………