Evaluate: ∫ 4 10 2 + 5 3 x - 2 d x .

Let I = ∫ 4 10 2 + 5 3 x - 2 d x ⇒ I = 2 ∫ 4 10 1 · d x + ∫ 4 10 5 3 x - 2 d x Using: ∫ 1 a x + b d x = 1 a ln ( a x + b ) + c , a x + b > 0 and ∫ a · d x = a x + C , a ∈ R Therefore, I = 2 x + 5 3 ln 3 x - 2 4 10 , valid for 3 x - 2 > 0 ⇒ I = 2 10 + 5 3 ln 3 10 - 2 - 2 4 + 5 3 ln 3 4 - 2 ⇒ I = 20 + 5 3 ln 28 - 8 + 5 3 ln 10 ⇒ I = 12 + 5 3 ln 28 - 5 3 ln 10 ⇒ I = 12 + 5 3 ln 28 10 ⇒ I = 12 + 5 3 ln 14 5 Hence, ∫ 4 10 2 + 5 3 x - 2 d x = 12 + 5 3 ln 14 5 .

Evaluate: ∫ 1 3 3 x - 4 2 x - 1 d x .

Let I = ∫ 1 3 3 x - 4 2 x - 1 d x ⇒ I = ∫ 1 3 3 x d x - ∫ 1 3 4 2 x - 1 d x Using: ∫ 1 a x + b d x = 1 a ln ( a x + b ) + c , a x + b > 0 Therefore, I = 3 ln x - 4 2 ln 2 x - 1 1 3 , valid for 2 x - 1 > 0 and x > 0 ⇒ I = 3 ln 3 - 2 ln 2 3 - 1 - 3 ln 1 - 2 ln 2 1 - 1 ⇒ I = ln 3 3 - ln 5 2 ⇒ I = ln 27 - ln 25 ⇒ I = ln 27 25 Hence, ∫ 1 3 3 x - 4 2 x - 1 d x = ln 27 25 .

Evaluate: ∫ - 1 3 2 x - 1 + 4 2 x + 3 d x .

Let I = ∫ - 1 3 2 x - 1 + 4 2 x + 3 d x ⇒ I = 2 ∫ - 1 3 x d x - ∫ - 1 3 1 · d x + ∫ - 1 3 4 2 x + 3 d x Using: ∫ 1 a x + b d x = 1 a ln ( a x + b ) + c , a x + b > 0 and ∫ x n d x = x n + 1 n + 1 + c , n ≠ - 1 Therefore, I = 2 × x 2 2 - x + 4 2 ln 2 x + 3 - 1 3 , valid for 2 x + 3 > 0 ⇒ I = x 2 - x + 2 ln 2 x + 3 - 1 3 ⇒ I = 3 2 - 3 + 2 ln 2 3 + 3 - - 1 2 - - 1 + 2 ln 2 - 1 + 3 ⇒ I = 6 + 2 ln 9 - 2 + 2 ln 1 ⇒ I = 4 + ln 9 2 ⇒ I = 4 + ln 81 Hence, ∫ - 1 3 2 x - 1 + 4 2 x + 3 d x = 4 + ln 81 .

Evaluate: ∫ 1 5 4 x 2 x - 1 d x .

Let I = ∫ 1 5 4 x 2 x - 1 d x ⇒ I = ∫ 1 5 2 2 x - 1 + 2 2 x - 1 d x ⇒ I = ∫ 1 5 2 + 2 2 x - 1 d x Using: ∫ 1 a x + b d x = 1 a ln ( a x + b ) + c , a x + b > 0 Therefore, I = 2 x + 2 2 ln 2 x - 1 1 5 , valid for 2 x - 1 > 0 ⇒ I = 2 5 + ln ( 2 5 - 1 ) - 2 1 + ln ( 2 1 - 1 ) ⇒ I = 10 + ln 9 - 2 + ln ( 1 ) ⇒ I = 8 + ln 9 Hence, ∫ 1 5 4 x 2 x - 1 d x = 8 + ln 9 .

Show that: ∫ 0 1 6 x 2 + 5 x 2 x - 5 d x = 23 2 - 25 ln 5 3 .

We need to show that ∫ 0 1 6 x 2 + 5 x 2 x - 5 d x = 23 2 - 25 ln 5 3 Let I = ∫ 0 1 6 x 2 + 5 x 2 x - 5 d x ⇒ I = ∫ 0 1 3 x + 10 + 50 2 x - 5 d x [Dividing numerator with denominator, then we get remainder as 50 and quotient as 3 x + 10 [ Using: ∫ 1 a x + b d x = 1 a ln ( a x + b ) + c Therefore, I = 3 × x 2 2 + 10 x + 50 2 ln 2 x - 5 0 1 ⇒ I = 3 × 1 2 + 10 + 25 ln 2 1 - 5 - 3 × 0 + 0 + 25 ln 2 0 - 5 ⇒ I = 3 2 + 10 + 25 ln - 3 - 25 ln - 5 ⇒ I = 23 2 - 25 ln 5 3 Hence, ∫ 0 1 6 x 2 + 5 x 2 x - 5 d x = 23 2 - 25 ln 5 3 .

The diagram shows part of the curve y = 2 x + 3 . Given that the shaded region has area 4 . find the value of k .

Given that, we have, Using: The area included between the curve y = f x , x - axis and ordinates x = a ; x = b is given by ∫ a b y d x = ∫ a b f x d x . Therefore, required area A = ∫ 1 k y d x = ∫ 1 k 2 x + 3 d x ⇒ 4 = 2 ∫ 1 k 1 x + 3 d x [Given that, area = 4 ] ⇒ 2 = ln x + 3 1 k ⇒ 2 = ln k + 3 - ln 1 + 3 ⇒ 2 = ln k + 3 - ln 4 ⇒ 2 = ln k + 3 4 ⇒ k + 3 4 = e 2 ⇒ k + 3 = 4 e 2 ⇒ k = 4 e 2 - 3 Hence, the required value of k = 4 e 2 - 3 .

The points P ( 1 , - 2 ) and Q ( 2 , k ) lie on the curve for which d y d x = 3 - 2 x . The tangents to the curve at the points P and Q intersect at the point R . Find the coordinates of R .

Given that, we have: d y d x = 3 - 2 x Apply integration on both sides, then we get: y = 3 x - 3 ln x + c [Since, integration is inverse process of differentiation] Therefore, y = 3 x - 3 ln x + c And also, given that, the points P ( 1 , - 2 ) and Q ( 2 , k ) lie on the curve y = 3 x - 3 ln x + c i.e., - 2 = 3 1 - 3 ln 1 + c [The point P 1 , - 2 lie on the curve] ⇒ - 2 = 3 + c ⇒ c = - 5 Thus, the equation of the curve is y = 3 x - 3 ln x - 5 . If point Q 2 , k is lie on the curve i.e., k = 3 2 - 3 ln 2 - 5 ⇒ k = 1 - ln 2 3 ⇒ k = 1 - ln 8 Now, the slope of the tangent at P 1 , - 2 is d y d x x = 1 = 3 - 2 1 = 1 So, the equation of the tangent at P 1 , - 2 is y - - 2 = x - 1 ⇒ y + 2 = x - 1 ⇒ x - y - 2 - 1 = 0 ⇒ x - y - 3 = 0 → 1 Now, the slope of the tangent at Q 2 , 1 - ln 8 is d y d x x = 2 = 3 - 2 2 = 2 So, the equation of the tangent at Q 2 , 1 - ln 8 is y - 1 - ln 8 = 2 x - 2 ⇒ y - 1 + ln 8 = 2 x - 4 ⇒ 2 x - y + 1 - ln 8 - 4 = 0 ⇒ 2 x - y - ln 8 - 3 = 0 → 2 And also, given that, the tangents 1 and 2 intersect at the point R . Now, 2 - 1 , then we get: x - ln 8 = 0 ⇒ x = ln 8 Substitute the value of x = ln 8 in 1 , then ln 8 - y - 3 = 0 ⇒ y = ln 8 - 3 Hence, the required co-ordinates of R = x , y = ln 8 , ln 8 - 3 .

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Integration, Exercise 3: EXERCISE 5B

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Integration, Exercise 3: EXERCISE 5B

Attempt the free practice questions on Chapter 5: Integration, Exercise 3: EXERCISE 5B with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Integration, Exercise 3: EXERCISE 5B with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 3

Evaluate: $\int_{4}^{10} (2 + \frac{5}{3 x - 2}) d x$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 3

Evaluate: $\int_{1}^{3} (\frac{3}{x} - \frac{4}{2 x - 1}) d x$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 3

Evaluate: $\int_{- 1}^{3} (2 x - 1 + \frac{4}{2 x + 3}) d x$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 4

Evaluate: $\int_{4}^{10} (2 + \frac{5}{3 x - 2}) d x = 12 + \frac{5}{3} \ln (\frac{14}{5})$ .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 5

Show that: $\int_{0}^{1} \frac{6 x^{2} + 5 x}{2 x - 5} d x = \frac{23}{2} - 25 \ln (\frac{5}{3})$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 6

A curve is such that $\frac{d y}{d x} = 2 x + \frac{3}{x + e}$ . Given that the curve passes through the point $(e, e^{2})$ find the equation of the curve.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 7

Question Image

The diagram shows part of the curve $y = \frac{2}{x + 3}$ . Given that the shaded region has area $4$ . find the value of $k$ .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 5 - Integration>EXERCISE 5B>Q 8

The points $P (1, - 2) and Q (2, k)$ lie on the curve for which $\frac{d y}{d x} = 3 - \frac{2}{x}$ . The tangents to the curve at the points $P and Q$ intersect at the point $R$ . Find the coordinates of $R$ .

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