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

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) \(\int_0^2\) e^x dx (NCERT)
(ii) \(\int_{-1}^5\) e^x 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) = e^x ;
nh = b – a = 2 – 0 = 2
Thus, \(\int_0^2\) e^x dx = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h [f(0) + f(h) + f(2h) + ………… + f(n – 1) h]

(ii) Comparing \(\int_{-1}^5\) e^x dx with \(\int_a^b\) f(x) dx
Here, f(x) = e^x ;
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 + e^x) dx
Solution:
Comparing \(\int_1^3\) (x² + 3x + e^x) dx with \(\int_a^b\) f(x) dx f(x) dx, we have
f(x) = x² + 3x + e^x; 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) + e^{1 + h}
f(a + 2h) = f(1 + 2h)
= (1 + 2h)² + 3(1 + 2h) + e^{1 + 2h}
…………………………
…………………………