The diagram shows a shaded region, bounded by the curves y = 2 x , y = - 2 , x = 1 and x = 4 . The area of the region can be written in the form p ln ( 2 ) + q . Find the values of p and of q .

Given curves are: y = 2 x , y = - 2 , x = 1 Now, the definite integral for the area between the two functions is A = ∫ x 1 x 2 g x - f x d x i.e., A = ∫ 1 4 2 x - ( - 2 ) d x Therefore, A = ∫ 1 4 2 x + 2 d x ⇒ A = 2 ln x + 2 x 1 4 ⇒ A = 2 ln 4 + 2 4 - 2 ln 1 + 2 1 ⇒ A = 2 ln 4 + 8 - 2 ⇒ A = 2 ln 2 2 + 6 ⇒ A = 4 ln 2 + 6 On comparing with the given expression p ln 2 + q , we have: p = 4 , q = 6 .

The diagram shows a shaded region, bounded by the curves y = 2 x , y = - 2 , x = 1 and x = 4 . The area of the region can be written in the form p ln ( 2 ) + q . Find the values of p and of q .

Given curves are: y = 2 x , y = - 2 , x = 1 Now, the definite integral for the area between the two functions is A = ∫ x 1 x 2 g x - f x d x i.e., A = ∫ 1 4 2 x - ( - 2 ) d x Therefore, A = ∫ 1 4 2 x + 2 d x ⇒ A = 2 ln x + 2 x 1 4 ⇒ A = 2 ln 4 + 2 4 - 2 ln 1 + 2 1 ⇒ A = 2 ln 4 + 8 - 2 ⇒ A = 2 ln 2 2 + 6 ⇒ A = 4 ln 2 + 6 On comparing with the given expression p ln 2 + q , we have: p = 4 , q = 6 .

Consider the functions g x = 1 2 x and f x = x . Part of the graphs of f and g are shown in the diagram. Find the shaded region area.

Given functions: g x = 1 2 x and f x = x Finding the x - coordinates of the points of intersection of the two functions. Now, the point of intersection of two curves can be calculated by equating f x = g x i.e., 1 2 x = x Therefore, x = 2 x ⇒ x 2 = 4 x [Squaring on both sides] ⇒ x 2 - 4 x = 0 ⇒ x x - 4 = 0 ⇒ x = 0 or 4 Now, if x = 0 , then we have y = 0 and similarly, if x = 4 , then we have y = 2 Hence, the coordinates of the point of intersection of graphs of f and g are: ( 0 , 0 ) , ( 4 , 2 ) . Now, the definite integral for the area between the two functions is A = ∫ x 1 x 2 ( f ( x ) - g ( x ) ) d x ⇒ A = ∫ 0 4 x - 1 2 x d x ⇒ A = ∫ 0 4 x 1 2 - 1 2 x d x ⇒ A = x 1 2 + 1 1 2 + 1 - 1 2 × x 1 + 1 1 + 1 0 4 ⇒ A = 2 3 x 3 2 - 1 4 x 2 0 4 ⇒ A = 2 3 4 3 2 - 1 4 4 2 - 0 ⇒ A = 16 3 - 4 ⇒ A = 4 3 sq.units Hence, the required area of the shaded reign is 4 3 sq.units.

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The diagram shows a shaded region, bounded by the curves $y = \frac{2}{x}, y = - 2, x = 1$ and $x = 4$ . The area of the region can be written in the form $p \ln (2) + q$ . Find the values of $p and of q$ .

Important Questions on From Approximation to Generalization: Integration

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. Find the coordinates of the points of intersection of the graphs of $f and g$ .

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. Write an integral expression for the area enclosed by the graph of $f and g$ .

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. Find the shaded region area.

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. The line $x = k$ divides the area of the region in half. Write down an integral expression for half the area of the region.

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. The line $x = k$ divides the area of the region in half. Find an expression, not involving an integral, for half the area of the region.

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10J>Q 10

Consider the functions $g (x) = \frac{1}{2} x and f (x) = \sqrt{x}$ . Part of the graphs of $f and g$ are shown in the diagram. The line $x = k$ divides the area of the region in half. Using your GDC, find the value of $k$ .

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10K>Q 1

Write an integral expression to find the area of the region bounded by the two curves $f (x) = x^{2} - 2 x and g (x) = 10 x + x^{2} - 3 x^{3}$ , and then find this area. Use your GDC to examine the graphs and evaluate any definite integrals.

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Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 10 - From Approximation to Generalization: Integration>Exercise 10K>Q 2

Write an integral expression to find the area of the region bounded by the two curves $f (x) = x^{3} - 3 x^{2} + 3 x + 1 and g (x) = x + 1$ , and then find this area. Use your GDC to examine the graphs and evaluate any definite integrals.

The diagram shows a shaded region, bounded by the curves y=2x,y=-2,x=1 and x=4. The area of the region can be written in the form pln(2)+q. Find the values of p and of q.

Important Questions on From Approximation to Generalization: Integration

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The diagram shows a shaded region, bounded by the curves $y = \frac{2}{x}, y = - 2, x = 1$ and $x = 4$ . The area of the region can be written in the form $p \ln (2) + q$ . Find the values of $p and of q$ .