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Question 1. Find the area under the given curves and given lines:(i) y = x2, x = 1, x = 2 and x-axis(ii) y = x4, x = 1, x = 5 and x-axisSolution: (i) For the curve ???? = ????2 between ???? = 1 and x = 2 and the x-axis: ![]() Area EFGH = ∫12 x2 dx Area = [ x3/3 ]12 Area = (23 /3) – (13 / 3) = 8/3 – 1/3 = 7/3 Areas under the curves and lines are: 7/3 square units (ii) For the curve y = x4 between x = 1 and x = 5 and the x-axis: ![]() Area EFGH = ∫15 x4 dx Area = [ x5/ 5]15 Area = (55/ 5) – (15/ 5) = 3125/5 = 1/5 = 624.8 The areas under the curves and lines are: 624.8 square units Question 2. Sketch the graph of y = |x + 3| and evaluate ∫-60 |x+3| dx.Solution: Given equation is y = |x + 3| Corresponding values of x and y are given in the following table.
On plotting these points, we obtain the graph of y = |x + 3| as follows It is know that , (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0∴ ∫-60 |(x+3)|dx = – ∫-6-3 (x+3)dx + ∫-30 (x+3)dx= -[x2 / 2 + 3x]-6-3 + [x2 / 2 + 3x]-30= -[((-3)2/2 + 3(-3))-((-6)2/2 + 3(-6)] + [0-((-3)2 + 3(-3))]= -[-9/2]-[-9/2]= 9 square units Question 3. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.Solution: Graph of y = sin x can be drawn as: Required area = Area OEFO + Area FGHF = ∫0π sinx dx + | ∫π 2π sin x dx| = [-cos x]0π + |[-cosx]x2π| = [-cosπ+cos0] + |-cos 2π + cosπ] = 1+1+|(-1-1)| = 2 + |-2| = 2+ 2 = 4 square units Question 4. Area bounded by the curve y = x3 , the x-axis and the ordinates x = – 2 and x = 1 is(A) – 9 (B) -15/4 (C) 15/4 (D) 17/4Solution: To find the area bounded by the curve ????= ????3 ,the x-axis, and the ordinates ???? = −2 and ????=1, we integrate x3 with respect to x over the interval [−2, 1]. Area= ∫1-2 x3dx Using the antiderivative of x3 which is x4/4, we have: Area = [x4/4]1-2 = (14/4)- ((-2)4 /4) = (1/4) – (16/4) = 1/4- 4/1 = -15/4 square units Option B is correct. Question 5. The area bounded by the curve y = x | x | , x-axis and the ordinates x = – 1 and x = 1 is given by(A) 0 (B) 1/ 3 (C) 2/3 (D) 4/3Solution: Curve y = x.∣x∣ has different definitions for positive and negative values of x: For x ≥ 0, |x| = x, so y = x2 For x < 0, |x| = -x, so y = -x2 We can integrate each part separately over their respective intervals: For x ≥ 0; Area1 = ∫01 x2 dx For x < 0; Area2 = ∫-10 -x2 dx Then, the total area is sum of these two areas: Total Area = Area1 + Area2 Let’s calculate: For x ≥ 0: Area1 = ∫01 x2 dx = [x3 / 3]01 = 1/3 For x < 0: Area2 = ∫-10 -x2 dx = [-x3/3]10 = 1/3 such that total area is : Total Area = Area1 + Area2 = 1/3 + 1/3 = 2/3 so, that correct option is 2/3 square units Related Articles: |
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Class 12 |
Type: | Geek |
Category: | Coding |
Sub Category: | Tutorial |
Uploaded by: | Admin |
Views: | 15 |