Consider the function f where f x = x x 2 - 1 , x ∈ ℝ . Determine an expression for the derivative of f .

Given f x = x x 2 - 1 , x ∈ ℝ . ⇒ f x = x 3 - x Differentiate with respect to x , we know that d d x x n = n x n - 1 Therefore, f ' x = 3 x 2 - 1

Consider the function f where f x = x x 2 - 1 , x ∈ ℝ . Determine an expression for the derivative of f .

Given f x = x x 2 - 1 , x ∈ ℝ . ⇒ f x = x 3 - x Differentiate with respect to x , we know that d d x x n = n x n - 1 Therefore, f ' x = 3 x 2 - 1

Consider the function f where f x = x x 2 - 1 , x ∈ ℝ . Hence find the x -coordinates of the turning points on the graph of f .

Given f x = x x 2 - 1 , x ∈ ℝ . ⇒ f x = x 3 - x Differentiate with respect to x , we know that d d x x n = n x n - 1 Therefore, f ' x = 3 x 2 - 1 To identify the turning on the graph, Consider f ' x = 0 ⇒ 3 x 2 - 1 = 0 ⇒ x = ± 1 3 or ⇒ x = - 1 3 , 1 3

Consider the function f where f x = x x 2 - 1 , x ∈ ℝ .. Explain why the value of the integral ∫ - 1 1 f ( x ) d x does not represent the area of the region S enclosed by the graph of f and the x -axis between x = ± 1

Given f x = x x 2 - 1 , x ∈ ℝ . Let I = ∫ - 1 1 f ( x ) d x We know that ∫ - a a f x d x = 2 ∫ 0 a f x d x i f f - x = f x 0 i f f - x = - f x Consider f - x = - x - x 2 - 1 = - x x 2 - 1 = - f x Therefore I = ∫ - 1 1 f ( x ) d x = 0 The function changes sign in the interval [ - 1 , 1 ] , so the areas above and below the x -axis cancel out.

Given x - 2 4 We know that ( a - b ) 4 = a 4 - 4 a 3 b + 6 a 2 b 2 - 4 a b 3 + b 4 Therefore, x - 2 4 = x 4 - 4 x 3 2 + 6 x 2 2 2 - 4 x 2 3 + 2 4 x - 2 4 = x 4 - 8 x 3 + 24 x 2 - 32 x + 16

Find ∫ x - 2 4 d x .

Let I = ∫ x - 2 4 d x We know that ( a - b ) 4 = a 4 - 4 a 3 b + 6 a 2 b 2 - 4 a b 3 + b 4 Therefore, x - 2 4 = x 4 - 4 x 3 2 + 6 x 2 2 2 - 4 x 2 3 + 2 4 x - 2 4 = x 4 - 8 x 3 + 24 x 2 - 32 x + 16 Therefore, I = ∫ x 4 - 8 x 3 + 24 x 2 - 32 x + 16 d x I = x 5 5 - 2 x 4 + 8 x 3 - 16 x 2 + 16 x + C

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 13 P2:

EASY

Diploma

IMPORTANT

Earn 100

Consider the function $f$ where $f (x) = x (x^{2} - 1)$ , $x \in ℝ$ .

Determine an expression for the derivative of $f$ .

Important Questions on From Approximation to Generalization: Integration

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 13 P2:

Consider the function $f$ where $f (x) = x (x^{2} - 1)$ , $x \in ℝ$ .

Hence find the $x$ -coordinates of the turning points on the graph of $f$ .

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 13 P2:

Consider the function $f$ where $f (x) = x (x^{2} - 1)$ , $x \in ℝ$ .

Find the value of $\int_{- 1}^{1} f (x) d x$ .

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 13 P2:

Consider the function $f$ where $f (x) = x (x^{2} - 1)$ , $x \in ℝ$ ..

Question Image

Explain why the value of the integral $\int_{- 1}^{1} f (x) d x$ does not represent the area of the region $S$ enclosed by the graph of $f$ and the $x$ -axis between $x = \pm 1$

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 13 P2:

Consider the function $f$ where $f (x) = x (x^{2} - 1)$ , $x \in ℝ$ ..

Question Image

Find the area of the region $S$ enclosed by the graph of $f$ and the $x$ -axis between $x = \pm 1$

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Chapter review>Q 14 P1:

Consider the functions $f$ and $g$ defined by $f (x) = x^{2}$ and $g (x) = 3 - 2 x$ . Show that the graphs of $f$ and $g$ intersect at points with $x$ - coordinates $- 3$ and $1$ .