What is the area of the region bounded by curve y=(x-1) ^ 2, y=(x+1) ^2, and y=1/4?
between the red, black and green lines
Sketch the region bounded by the given equations.?
Sketch the region bounded by the given equations. Express the region's area as an iterated double integral and evaluate the integral. y=e^x, y=0, x=0, x=ln12 What is the correct form of the integral? What is the area of the region_____?
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 the given curves. y = 15 − x2, y = x2 − 3?
at a guess, this is actually y = 15 − x² y = x² − 3 otherwise there is no enclosed area. Please use proper symbols. find the intersections, at about x = ±3, y = 6, by solving the two equations. the curves are symmetrical, and if you moved the top one down by y = 6, to y = 9 – x² and the x axis becomes the red line, then the area is twice the integral of that from –3 to +3 Thus the area is 2∫ [–3, +3] 9 – x² dx or 2 [–3, +3] 9x – (x³/3) or 2( [27 – 27/3] – [–27 + 27/4] ) or 2 (27 + 27 – 9 – 9 ) = 72
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find?
Picture: http://a.imageshack.us/img713/7506/91582... A = ∫ ∫ dx dy {(x,y)| 2-y/2 ≤ x ≤ y²/4 ; 2 ≤ y ≤ 3} = ∫ y²/4 - (2 - y/2) dy {(y)| 2 ≤ y ≤ 3} = 5/6 Answer: 5/6 units²
What is the area bounded by x-axis and the curve [math]y = 4x-x^2[/math]?
First, we will find the area of the region bounded by the curve and x-axis.[math]\because[/math] At x-axis, [math]y = 0[/math]So, we find the area between:[math]y = 4x - x^2[/math] … (i)[math]y = 0[/math]To determine the area of the shaded region between curve and x-axis, we need to sketch this curve on the graph.Graphical representation:Here, shaded region represents the area bounded by [math]y = 4x - x^2[/math] and [math]y = 0[/math]Area of the shaded region:Let [math]A[/math] be the area of the shaded region.Then[math]\boxed {A = \displaystyle \int_a^b [ f(x) - g(x)] \;dx}[/math]Here, [math]f(x)[/math] is the top curve and [math]g(x)[/math] is the bottom curve.[math]a[/math] and [math]b[/math] are the limits as[math]a[/math] (Lower limit) = [math]x[/math] coordinate of extreme left intersection point of area to be found.[math]b[/math] (Upper limit) = [math]x[/math] coordinate of extreme right intersection point of area to be found.So, [math]f(x) = y = 4x - x^2[/math][math]g(x) = y = 0[/math]We need to find the limits, [math]a[/math] and [math]b[/math]How to Find the limits:Since, limits [math]a[/math] and [math]b[/math] are the [math]x[/math] coordinates of the intersection points of the given curves.[math]\therefore\;[/math] We will find the intersection points of the given curve and line.Put the value of [math]y[/math] from equation (ii) into equation (i)[math]4x - x^2 = 0[/math][math]x (4-x) = 0[/math][math]x = 0,\;x=4[/math]Put these values in equation (i)[math]y = 0, \;y = 0[/math]Thus the points of intersection are [math]O(0,0)[/math] and [math]A(4, 0)[/math][math]\therefore, \;a = 0, b = 4[/math]Area between curves:[math]A = \displaystyle \int_a^b [ f(x) - g(x)] \;dx =\displaystyle \int_0^4 [ (4x-x^2) - 0] \;dx[/math][math]=\left [ 4\left( \dfrac {x^2}{2}-\dfrac {x^3}{3} \right) \right]=\left [ 32-\dfrac {64}{3}\right] - 0[/math][math]=\left [ \dfrac {96-64}{3}\right][/math][math]A=\dfrac {32}{3}[/math][math] [/math]
Find the area of the region enclosed by the given curves?
The equations are: y = 6 cos 5x, y = 6 sin 10x, x = 0, x = π/10 My algebra is probably the thing that's messing me up, and I can't find where 6cos5x=6sin10x in order to know where to place the integrals in the formulas. If you could show steps so I could see where I went wrong, it would be greatly appreciated!