Short Question 5 – 7

Posted on April 22, 2020 by user

Question 5:

\begin{array}{l} Given \int (6 x^{2} + 1) d x = m x^{3} + x + c, \\ where m and c are constants, find \\ (a) the value of m . \\ (b) the value of c if \int (6 x^{2} + 1) d x = 13 when x = 1. \end{array}

Solution:
(a)

\begin{array}{l} \int (6 x^{2} + 1) d x = m x^{3} + x + c \\ \frac{6 x^{3}}{3} + x + c = m x^{3} + x + c \\ 2 x^{3} + x + c = m x^{3} + x + c \\ Compare the both sides, \\ ∴ m = 2 \end{array}

(b)

\begin{array}{l} \int (6 x^{2} + 1) d x = 13 when x = 1. \\ 2 {(1)}^{3} + 1 + c = 13 \\ 3 + c = 13 \\ c = 10 \end{array}

Question 6:

\begin{array}{l} It is given that \int_{5}^{k} g (x) d x = 6, and \int_{5}^{k} [g (x) + 2] d x = 14, \\ find the value of k . \end{array}

Solution:

\begin{array}{l} \int_{5}^{k} [g (x) + 2] d x = 14 \\ \int_{5}^{k} g (x) d x + \int_{5}^{k} 2 d x = 14 \\ 6 + {[2 x]}_{5}^{k} = 14 \\ 2 (k - 5) = 8 \\ k - 5 = 4 \\ k = 9 \end{array}

Question 7:

Given \int_{k}^{2} (4 x + 7) d x = 28, calculate the possible value of k .

Solution:

\begin{array}{l} \int_{k}^{2} (4 x + 7) d x = 28 \\ {[2 x^{2} + 7 x]}_{k}^{2} = 28 \\ 8 + 14 - (2 k^{2} + 7 k) = 28 \\ 22 - 2 k^{2} - 7 k = 28 \\ 2 k^{2} + 7 k + 6 = 0 \\ (2 k + 3) (k + 2) = 0 \\ k = - \frac{3}{2} or k = - 2 \end{array}

Short Question 8 – 10

Posted on April 22, 2020 by user

Question 8:

Given y = \frac{5 x}{x^{2} + 1} and \frac{d y}{d x} = g (x), find the value of \int_{0}^{3} 2 g (x) d x .

Solution:

\begin{array}{l} Since \frac{d y}{d x} = g (x), thus y = \int g (x) d x \\ \int_{0}^{3} 2 g (x) d x = 2 \int_{0}^{3} g (x) d x \\ = 2 {[y]}_{0}^{3} \\ = 2 {[\frac{5 x}{x^{2} + 1}]}_{0}^{3} \\ = 2 [\frac{5 (3)}{3^{2} + 1} - 0] \\ = 2 (\frac{15}{10}) \\ = 3 \end{array}

Question 9:

Find \int_{5}^{k} (x + 1) d x, in terms of k .

Solution:

\begin{array}{l} {\int_{5}^{k} (x + 1) d x = [\frac{x^{2}}{2} + x]}_{5}^{k} \\ = (\frac{k^{2}}{2} + k) - (\frac{5^{2}}{2} + 5) \\ = \frac{k^{2} + 2 k}{2} - \frac{35}{2} \\ = \frac{k^{2} + 2 k - 35}{2} \end{array}

Question 10:

\begin{array}{l} Given that = \int_{2}^{5} g (x) d x = - 2 . Find \\ (a) the value of \int_{5}^{2} g (x) d x, \\ (b) the value of m if \int_{2}^{5} [g (x) + m (x)] d x = 19 \end{array}

Solution:

\begin{array}{l} (a) \int_{5}^{2} g (x) d x = - \int_{2}^{5} g (x) d x \\ = - (- 2) \\ = 2 \end{array}

\begin{array}{l} (b) \int_{2}^{5} [g (x) + m (x)] d x = 19 \\ \int_{2}^{5} g (x) d x + m \int_{2}^{5} x d x = 19 \\ - 2 + m {[\frac{x^{2}}{2}]}_{2}^{5} = 19 \\ \frac{m}{2} {[x^{2}]}_{2}^{5} = 21 \\ \frac{m}{2} [25 - 4] = 21 \\ 21 m = 42 \\ m = 2 \end{array}

Short Question 11 – 13

Posted on April 22, 2020 by user

Question 11:

\begin{array}{l} Given \int_{- 2}^{3} g (x) d x = 4, and \int_{- 2}^{3} h (x) d x = 9, find the value of \\ (a) \int_{- 2}^{3} 5 g (x) d x, \\ (b) m if \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \end{array}

Solution:
(a)

\begin{array}{l} \int_{- 2}^{3} 5 g (x) d x = 5 \int_{- 2}^{3} g (x) d x \\ = 5 \times 4 \\ = 20 \end{array}

(b)

\begin{array}{l} \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \\ \int_{- 2}^{3} g (x) d x + 3 \int_{- 2}^{3} h (x) d x + \int_{- 2}^{3} 4 m d x = 12 \\ 4 + 3 (9) + 4 m {[x]}_{- 2}^{3} = 12 \\ 4 m [3 - (- 2)] = - 19 \\ 20 m = - 19 \\ m = - \frac{19}{20} \end{array}

Question 12:

\begin{array}{l} (a) Find the value of \int_{- 1}^{1} {(3 x + 1)}^{3} d x . \\ (b) Evaluate \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x . \end{array}

Solution:

\begin{array}{l} a) {\int_{- 1}^{1} {(3 x + 1)}^{3} d x = [\frac{{(3 x + 1)}^{4}}{4 (3)}]}_{- 1}^{1} \\ = {[\frac{{(3 x + 1)}^{4}}{12}]}_{- 1}^{1} \\ = \frac{1}{12} [4^{4} - {(- 2)}^{4}] \\ = \frac{1}{12} (256 - 16) \\ = 20 \end{array}

\begin{array}{l} (b) \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x = \int_{3}^{4} \frac{1}{{(2 x - 4)}^{\frac{1}{2}}} d x \\ = \int_{3}^{4} {(2 x - 4)}^{- \frac{1}{2}} d x \\ = {[\frac{{(2 x - 4)}^{- \frac{1}{2} + 1}}{\frac{1}{2} (2)}]}_{3}^{4} \\ = {[\sqrt{2 x - 4}]}_{3}^{4} \\ = [\sqrt{2 (4) - 4} - \sqrt{2 (3) - 4}] \\ = 2 - \sqrt{2} \end{array}

Question 13:

\begin{array}{l} Given that y = \frac{x^{2}}{2 x - 1}, show that \\ \frac{d y}{d x} = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} . Hence, evaluate \int_{- 2}^{2} \frac{x (x - 1)}{4 {(2 x - 1)}^{2}} d x . \end{array}

Solution:

\begin{array}{l} y = \frac{x^{2}}{2 x - 1} \\ \frac{d y}{d x} = \frac{(2 x - 1) (2 x) - x (2)}{{(2 x - 1)}^{2}} \\ = \frac{4 x^{2} - 2 x - 2 x^{2}}{{(2 x - 1)}^{2}} \\ = \frac{2 x^{2} - 2 x}{{(2 x - 1)}^{2}} \\ = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} (shown) \\ \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{8} \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{4} \int_{- 2}^{2} \frac{x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} [(\frac{2^{2}}{2 (2) - 1}) - (\frac{{(- 2)}^{2}}{2 (- 2) - 1})] \\ = \frac{1}{8} [(\frac{4}{3}) - (\frac{4}{- 5})] \\ = \frac{1}{8} (\frac{32}{15}) \\ = \frac{4}{15} \end{array}

Long Question 3

Posted on April 22, 2020 by user

Question 3:

The gradient function of a curve which passes through P(2, –14) is 6x² – 12x.

Find

(a) the equation of the curve,

(b) the coordinates of the turning points of the curve and determine whether each of the turning points is a maximum or a minimum.

Solution:
(a)

\begin{array}{l} Given gradient function of a curve \frac{d y}{d x} = 6 x^{2} - 12 x \\ The equation of the curve, \\ y = \int (6 x^{2} - 12 x) d x \\ y = \frac{6 x^{3}}{3} - \frac{12 x^{2}}{2} + c \\ y = 2 x^{3} - 6 x^{2} + c \\ - 14 = 2 {(2)}^{3} - 6 {(2)}^{2} + c, at point P (2, - 14) \\ - 14 = - 8 + c \\ c = - 6 \\ y = 2 x^{3} - 6 x^{2} - 6 \end{array}

(b)

\begin{array}{l} \frac{d y}{d x} = 6 x^{2} - 12 x \\ At turning points, \frac{d y}{d x} = 0 \\ 6 x^{2} - 12 x = 0 \\ 6 x (x - 2) = 0 \\ x = 0, x = 2 \\ x = 0, y = 2 {(0)}^{3} - 6 {(0)}^{2} - 6 = - 6 \\ x = 2, y = 2 {(2)}^{3} - 6 {(2)}^{2} - 6 = - 14 \\ \frac{d^{2} y}{d x^{2}} = 12 x - 12 \\ When x = 0, \frac{d^{2} y}{d x^{2}} = 12 (0) - 12 = - 12 < 0 \\ (0, - 6) is a maximum point . \\ When x = 2, \frac{d^{2} y}{d x^{2}} = 12 (2) - 12 = 12 > 0 \\ (2, - 14) is a minimum point . \end{array}

3.5 Integration as the Summation of Areas

Posted on April 22, 2020 by user

3.5 Integration as the Summation of Areas

(A) Area of the region between a Curve and the x-axis.

Area of the shaded region;

A = \int_{a}^{b} y d x

(B) Area of the region between a curve and the y-axis.

Area of the shaded region;

A = \int_{a}^{b} x d y

(C) Area of the region between a curve and a straight line.

Area of the shaded region;

A = \int_{a}^{b} f (x) d x - \int_{a}^{b} g (x) d x

3.2 Integration by Substitution

Posted on April 22, 2020 by user

3.2 Integration by Substitution

It is given that

{\int (a x + b)}^{n} d x, n \neq - 1.

(A) Using the Substitution method,

\begin{array}{l} Let u = a x + b \\ Thus, \frac{d u}{d x} = a \\ ∴ d x = \frac{d u}{a} \end{array}

Example 1:

\begin{array}{l} {\int (3 x + 5)}^{3} d x . \\ Let u = 3 x + 5 \\ \frac{d u}{d x} = 3 \\ d x = \frac{d u}{3} \\ {\int (3 x + 5)}^{3} d x \\ = \int u^{3} \frac{d u}{3} \leftarrow \begin{array}{l} substitute 3 x + 5 = u \\ and d x = \frac{d u}{3} \end{array} \\ = \frac{1}{3} \int u^{3} d u \\ = \frac{1}{3} (\frac{u^{4}}{4}) + c \\ = \frac{1}{3} (\frac{{(3 x + 5)}^{4}}{4}) + c \leftarrow \begin{array}{l} substitute back \\ u = 3 x + 5 \end{array} \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}

(B) Using Formula method

\begin{array}{l} \int {(a x + b)}^{n} = \frac{{(a x + b)}^{n + 1}}{(n + 1) a} + c \\ Hence, \\ \int {(3 x + 5)}^{3} d x = \frac{{(3 x + 5)}^{4}}{4 (3)} + c \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}

Example 2 (Formula method):

\begin{array}{l} \int {(a x + b)}^{n} = \frac{{(a x + b)}^{n + 1}}{(n + 1) a} + c \\ Hence, \\ \int {(3 x + 5)}^{3} d x = \frac{{(3 x + 5)}^{4}}{4 (3)} + c \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}

Basic Integration

Posted on April 22, 2020 by user

3.1 Integration as the Inverse of Differentiation, Integration of axⁿ and integration of the Functions of the Sum/Difference of Algebraic Terms

Type 1:

\begin{array}{l} \int a d x = a x + C \\ Example \\ \int 2 d x = 2 x + C \end{array}

Type 2:

\begin{array}{l} \int a x^{n} d x = \frac{a x^{n + 1}}{n + 1} + C \\ E x a m p l e 1 \\ \int 2 x^{3} d x = \frac{2 x^{4}}{4} + C = \frac{x^{4}}{2} + C \\ E x a m p l e 2 \\ \int \frac{2}{3 x^{5}} d x = \int \frac{2}{3} x^{- 5} d x = \frac{2}{3} (\frac{x^{- 4}}{- 4}) + C \\ = \frac{2}{3} (\frac{x^{- 4}}{- 4}) + C = \frac{x^{- 4}}{- 6} + C \end{array}

Type 3:

\begin{array}{l} \int (u + v) d x = \int u d x \pm \int v d x \\ u and v are functions in x \\ E x a m p l e 1 \\ \int 3 x^{2} + 2 x d x = \int 3 x^{2} d x + \int 2 x d x \\ = \frac{3 x^{3}}{3} + \frac{2 x^{2}}{2} + C \\ = \frac{3 x^{3}}{3} + \frac{2 x^{2}}{2} + C \\ = x^{3} + x^{2} + C \end{array}

\begin{array}{l} E x a m p l e 2 \\ \int (x + 2) (3 x + 1) d x = \int 3 x^{2} + 7 x + 2 d x \\ = \int 3 x^{2} d x + \int 7 x d x + \int 2 d x \\ = \frac{3 x^{3}}{3} + \frac{7 x^{2}}{2} + 2 x + C \\ = x^{3} + \frac{7 x^{2}}{2} + 2 x + C \end{array}

\begin{array}{l} E x a m p l e 3 \\ \int \frac{3 x^{3} + x^{2} - x}{x} d x = \int 3 x^{2} + x - 1 d x \\ = \int 3 x^{2} d x + \int x d x - \int 1 d x \\ = \frac{3 x^{3}}{3} + \frac{x^{2}}{2} - x + C \\ = x^{3} + \frac{x^{2}}{2} - x + C \end{array}

Finding Equation Of A Curve From Its Gradient Function

Posted on April 22, 2020 by user

3.3 Finding Equation of a Curve from its Gradient Function

Example 1:
Find the equation of the curve that has the gradient function

\frac{d y}{d x} = 2 x + 8

and passes through the point (2, 3).

Solution:

\begin{array}{l} y = \int (2 x + 8) \\ y = \frac{2 x^{2}}{2} + 8 x + C \end{array}

y = x² + 8x + C

3 = 2² +8(2) + C (2, 3)

C = –17
Hence, the equation of the curve is
y = x² + 8x – 17

Example 2:

The gradient function of a curve is given by 2x – 4 and the curve has a minimum value of 3. Find the equation of the curve.

Solution:

At the point where a curve has a minimum value,

\frac{d y}{d x} = 0

\frac{d y}{d x} = 0

2x – 4 = 0

x = 2

Therefore minimum point = (2, 3).

\begin{array}{l} \frac{d y}{d x} = 2 x - 4 \\ y = \int (2 x - 4) d x \\ y = \frac{2 x^{2}}{2} - 4 x + C \\ y = x^{2} - 4 x + C \end{array}

When x = 2, y = 3.

3 = 2² – 4(2) + c

c = 7

Hence, the equation of the curve is

y = x² – 4x + 7

Definite Integrals

Posted on April 22, 2020 by user

3.4a Definite Integral of f(x) from x=a to x=b

Example:

Evaluate each of the following.

\begin{array}{l} (a) \int_{- 1}^{0} (3 x^{2} - 2 x + 5) d x \\ (b) \int_{0}^{2} {(2 x + 1)}^{3} d x \end{array}

Solution:

\begin{array}{l} (a) \int_{- 1}^{0} (3 x^{2} - 2 x + 5) d x \\ = {[\frac{3 x^{3}}{3} - \frac{2 x^{2}}{2} + 5 x]}_{- 1}^{0} \\ = {[x^{3} - x^{2} + 5 x]}_{- 1}^{0} \\ = 0 - [{(- 1)}^{3} - {(- 1)}^{2} + 5 (- 1)] \\ = 0 - (- 1 - 1 - 5) \\ = 7 \\ (b) \int_{0}^{2} {(2 x + 1)}^{3} d x \\ = {[\frac{{(2 x + 1)}^{4}}{4 (2)}]}_{0}^{2} \\ = {[\frac{{(2 x + 1)}^{4}}{8}]}_{0}^{2} \\ = [\frac{{(2 (2) + 1)}^{4}}{8}] - [\frac{{(2 (0) + 1)}^{4}}{8}] \\ = \frac{625}{8} - \frac{1}{8} \\ = 78 \end{array}

Short Question 2 – 4

Posted on April 22, 2020 by user

Question 2:

Given that

Given that \int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c,

find the values of m and n.

Solution:

\begin{array}{l} \int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c \\ \int 4 {(1 + x)}^{- 4} d x = m {(1 + x)}^{n} + c \\ \frac{4 {(1 + x)}^{- 3}}{- 3 (1)} + c = m {(1 + x)}^{n} + c \\ - \frac{4}{3} {(1 + x)}^{- 3} + c = m {(1 + x)}^{n} + c \\ m = - \frac{4}{3}, n = - 3 \end{array}

Question 3:

\begin{array}{l} Given \int_{- 1}^{2} 2 g (x) d x = 4, and \int_{- 1}^{2} [m x + 3 g (x)] d x = 15. \\ Find the value of constant m . \end{array}

Solution:

\begin{array}{l} \int_{- 1}^{2} [m x + 3 g (x)] d x = 15 \\ \int_{- 1}^{2} m x d x + \int_{- 1}^{2} 3 g (x) d x = 15 \\ {[\frac{m x^{2}}{2}]}_{- 1}^{2} + 3 \int_{- 1}^{2} g (x) d x = 15 \\ [\frac{m {(2)}^{2}}{2} - \frac{m {(- 1)}^{2}}{2}] + \frac{3}{2} \int_{- 1}^{2} 2 g (x) d x = 15 \\ 2 m - \frac{1}{2} m + \frac{3}{2} (4) = 15 \leftarrow given \int_{- 1}^{2} 2 g (x) d x = 4 \\ \frac{3}{2} m + 6 = 15 \\ \frac{3}{2} m = 9 \\ m = 9 \times \frac{2}{3} \\ m = 6 \end{array}

Question 4:

Given \frac{d}{d x} (\frac{2 x}{3 - x}) = g (x), find \int_{1}^{2} g (x) d x .

Solution:

\begin{array}{l} Given \frac{d}{d x} (\frac{2 x}{3 - x}) = g (x) \\ \int g (x) d x = \frac{2 x}{3 - x} \\ Thus, \\ \int_{1}^{2} g (x) d x = {[\frac{2 x}{3 - x}]}_{1}^{2} \\ = \frac{2 (2)}{3 - 2} - \frac{2 (1)}{3 - 1} \\ = 4 - 1 \\ = 3 \end{array}