Sketch the region enclosed by y=3x and y=5x^2. Decide whether to integrate with respect to x or y then find.?
Well, you do want to set the equations equal to each other and solve for x, because those solutions help you find the bounds of integration. You're not looking for max or min values. 3x = 5x² ................ [ this is clearly true when x = 0 , the trivial solution ] 3 = 5x .................. [ divided both sides by x ] (3/5) = x ............... [ divided both sides by 5 ] y = 3x is a line. y = 5x² is an upwards-opening parabola. The points of intersection are (0,0) and ( 3/5 , 9/5 ) . If you sketch the graphs, you can see that the line is above the parabola in the enclosed region, and integrating with respect to x will work just fine, and the area is 3/5 ∫ (3x - 5x²) dx 0 = [(3x² / 2) - (5x³ / 3)] ............. evaluated at x = 0 and x = 3/5 ** The evaluation gets a bit 'messy* thanks to the fractional value 3/5 that you have to put in. You work through it and then you have to get things over a common denominator at the end. You could integrate with respect to y in this problem as well, but you would still have a fractional value to deal with, so nothing would be gained by that. ** = [(3 * 9/25) / 2] - [(5 * 27/125) / 3] = [(27/25) / 2] - [(135/125) / 3] = (27/50) - (135/375) = (405/750) - (270/750) = (135/750) = (27/150) = (9/50) ~~~~~~~ I made a graph of the situation. It's the best I could do with the graph, but if you look carefully at the grid lines and notice the size of the unit square, and then look at the area enclosed by the two functions, you can see that the area enclosed is a little less than 20% of a unit square. The answer above is (9/50) = 0.18 , so the graph gives a visual confirmation of the correctness of the answer. http://s1164.photobucket.com/user/iago9/...
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typ?
Determine the y-coordinates of the intersection pts: 6y + y² = 7 y² + 6y - 7 = 0 (y - 1)(y + 7) = 0 y = {1,-7} Take a look at the graph here: http://www.wolframalpha.com/input/?i=6x+%2B+y%C2%B2+%3D+7+and+y+%3D+x+for+-7+%3C+x+%3C+1 Note that the right function is 6x + y² = 7 or x = (7 - y²)/6, and the left side function is y = x. Integrate the area of the difference of two expressions to get: A = int.(y = -7,1) (7 - y²)/6 - y dy = 128/9 I hope this helps!
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find?
first let's look at the points where they intersect 2x=5x^2 2x-5x^2=0 x(2-5x)=0 x = 0 and x =2/5 area = integral 0 to 2/5 2x-5x^2 dx = (x^2 -5x^3/3) 0 to 2/5 = 4/25-8/75 =4/75
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y?
I don't know how to send you a sketch, but I hope you've sketched these graphs already. x+y = 0 is a straight line through O inclined at -45° to the positive x axis; x + y² = 30 is a parabola with horizontal axis, and since the equation can be written as x = 30 - y² it has vertex at (30, 0) and opens up towards the left, crossing the y axis at (0, ±√30) The graphs cross where x² + x - 30 = 0 i.e. (x + 6)(x - 5) = 0 i.e. at (-6, 6) and (5, -5) Integrate with respect to x or y? Since it is more straightforward to make x the subject of the parabola equation (to make y the subject would involve ±√), it is best to have the function expressions in terms of y, i.e. use ∫x dy for the area. = ∫(30 - y² - (-y))dy = [30y - y³/3 + y²/2] from -5 to 6 = [180 - 72 + 18] - [-150 -(-125/3) + 25/2] = 126 + 95 and 5/6 = 221 and 5/6 i.e. 221.8333333.... Please check my arithmetic -- did it in my head when not fully sober!
Sketch the region enclosed by x + y^2 = 2 and x + y = 0 . Decide whether to integrate with respect to x or y.?
It is more convenient to integrate dy, because the region would have to be broken into two subregions in order to integrate dx. The integrand is: -y^2 + 2 + y with y-limits from -1 to 2 After integration, we have: -(1/3)y^3 + 2y + (1/2)y^2 The final answer is: 9/2
Sketch the region enclosed by x+y^2=56 and x+y=0. Decide whether to integrate with respect to x or y.?
x + y² = 56 is a parabola with the x axis as its axis of symmetry, its vertex at (56, 0) and cutting the y axis at (0, ±√56) The line x + y = 0 cuts the parabola at (-8, 8) and (7, -7) You should be able to sketch this easily for yourself, but can also see it at Wolfram Alpha: http://www.wolframalpha.com/input/?i=x%2... Integrating with respect to x requires splitting into two regions: x = -8 to x = 7, then x = 7 to x = 56 with a different equation for each. So instead, integrate with respect to y from -7 to 8. At level y, the horizontal line-segment from x + y = 0 to x + y² = 56 has length x₂ - x₁ = (56-y²) - (-y) ......... = 56 + y - y² Hence the area of the region is [56y + y²/2 - y³/3] from -7 to 8 = 448 + 32 - 512/3 - (-392 + 49/2 + 343/3) = 562.5 sq unit
HELP Sketch the region enclosed by y=e^(2x), y=e^(4x), x=1. Decide whether to integrate with respect to x or y?
y=e^2x and y=e^4x intersect for e^2x=e^4x e^2x(1-e^2x)=0 then e^2x = 0 no real x because e^2x>0 for any x or (1-e^2x)=0 then 2x=0 ; x=0 REspect to x 1 ∫(e^4x-e^2x) dx =1/4e^4x-1/e^2x |1 0_________________________ |0 1/4e^4-1/2e^2-1/4e^0+1/2e^0= 1/4e^4-1/2e^2-1/4
Sketch the bounded region enclosed by y=e^(5x),y=e^(7x) and x=1. Confused...so confused..pls help?
Region bounded by: y = e⁵ˣ, y = e⁷ˣ, and x = 1 https://imgur.com/HhdD9dO We can see that it is best to integrate with respect to x: A = ∫ [0 to 1] (e⁷ˣ − e⁵ˣ) dx A = (1/7 e⁷ˣ − 1/5 e⁵ˣ) | [0 to 1] A = (1/7 e⁷ − 1/5 e⁵) − (1/7 − 1/5) A = 1/35 (5e⁷ − 7e⁵ + 2) A = 127.03639 We could still integrate with respect to y, but we'd need 2 integrals: A = ∫ [1 to e⁵] (1/5 ln y − 1/7 ln y) dy + ∫ [e⁵ to e⁷] (1 − 1/7 ln y) dy A = ∫ [1 to e⁵] (2/35 ln y) dy + ∫ [e⁵ to e⁷] (1 − 1/7 ln y) dy A = 2/35 (y ln y − y) | [1 to e⁵] + (y − 1/7 (y ln y − y)) | [e⁵ to e⁷] A = 2/35 [(e⁵ ln(e⁵) − e⁵) − (0 − 1)] + [(e⁷ − 1/7 (e⁷ ln(e⁷) − e⁷)) − (e⁵ − 1/7 (e⁵ ln(e⁵) − e⁵))] A = 2/35 [(5e⁵ − e⁵) + 1] + 1/7 [(7e⁷ − (7e⁷ − e⁷)) − (7e⁵ − (5e⁵ − e⁵))] A = 2/35 (4e⁵ + 1) + 1/7 (e⁷ − 3e⁵) A = 1/35 (8e⁵ + 2 + 5e⁷ − 15e⁵) A = 1/35 (5e⁷ − 7e⁵ + 2)
Sketch the region in the first quadrant enclosed. Decide whether to integrate with respect to x or y. Then find the area of the region.?
y=7/x y=2x y=(1/2)x 7/x=2x 7=2x^2 x^2=3.5 x= sqrt(3.5) = 1.8708 y= 7/x and y=2x intersect at 1.8708 First part of the area = ∫[0,1.8708] (2x-x/2) dx = (3/4)x^2 F(x) = (3/4)x^2 F(1.8708) = 2.6249 F(0) = 0 Fist area = F(1.8708)-F(0) = 2.6249 The second part area = 7/x= x/2 x^2=14 x=sqrt(14) = 3.7417 Second area = ∫ [1.8708,3.7417] (7/x-x/2) dx ∫ (7/x-x/2) = 7 ln(x) - (1/4)x^2 F(x) = 7 ln(x) - (1/4)x^2 F(3.7412) = 5.7367 F(1.8708) = 3.5096 F(3.7417)-F(1.8708) = 2.2271 First area + second area = 2.6249+2.2271 = 4.852 (area) <---------- Integration with respect to x) The triangle is split as two triangles https://gyazo.com/4591aeb0d6fcba69b6eded...