Guide :

limit int x (x --> 1/2)

You can use substitution but idk how to get there

Research, Knowledge and Information :

lim(x-->1+) x^2-1/abs x-1 - search results -

... (x-->1+) x^2-1/abs x-1. ... (x-2) at a. x-1 math Describe all the integer values of x that satisfy abs(x) ... Find the limit if it exists. lim 1/(x-2) ...
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How to disprove that 'the limit of int(x) as x approaches to ...

How can I disprove that "the limit of int(x) as x approaches to 1" does not exist, using the delta epsilon proof?
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limit x->1 x*2 - 1 / (x-1)*2 -

limit x->1 - 2936076. 1. Log in Join now Katie; ... Limit x->1 x*2 - 1 / (x-1)*2 1. Ask for details ; Follow; Report; by kirkland 02/21/2017. Log in to add a comment
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Section 2.2: The Limit of a Function - Saylor - Saylor Academy

x - [x] = x - INT(x) Fig. 8 INT(x) cc x Fig. 6 Fig. 1 . Created Date: 20111118193420Z ...
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lim([-1/(x+2)]+1/2)/x as x->0 - page 8 -

... [-1/(x+2)]+1/2)/x as x->0. ... omega int'l college x sinx ... Which would be the best option for finding the limit? lim (1-2x-3x^2)/1+x x->-1 A.Direct ...
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Lim Int X As X Approaches 1- Hi, I Understnad What... |

Answer to lim int x as x approaches 1- Hi, ... i understnad what limits are and that the the negativepower means ... Get this answer with Chegg Study View this ...
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Visual Calculus - Limits & the Greatest Integer Function

for values of x near 2. [[ x ]] denotes the greatest integer less than or equal to x. ... int x. and press the ENTER key. int(x) is the greatest integer function on ...
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Solutions to Limits as x Approaches a Constant

Now the limit can be computed. ) ... (Divide out the factors x - 1 , the factors which are causing the indeterminate form . Now the limit can be computed. ) .
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Suggested Questions And Answer :

find definite integral of (x^2)(sin2x) dx on an interval [-1,1]

Before going into the calculus, consider the curve itself. Note that if f(x)=x^2sin2x, f(x)=-f(-x), because x^2 is always positive while sin2x=-sin(-2x). Thus, where f(x) for positive x is above the x axis, the point for the corresponding negative value of x is below the x axis by the same amount. And where positive x is below the x axis, the corresponding negative value is above the x axis by the same amount. When we integrate between limits -n and n to find the area under the graph we will always get zero. So we can expect the result between -1 and 1 to be zero. On with the calculus. Let u=x^2 then du/dx=2x; let dv=sin2xdx, so dv/dx=sin2x, then v=-cos2x/2. d(uv)/dx=vdu/dx+udv/dx, so udv/dx=d(uv)/dx-vdu/dx; thus x^2sin2xdx=d(-x^2cos2x/2)/dx-int(-cos2x/2.2xdx). Integrating:  int(x^2sin2xdx)=-x^2cos2x/2+int(xcos2xdx). Now, let u=x, dv=cos2xdx; du/dx=1; v=sin2x/2. Applying the same formula and integrating: int(xcos2xdx)=xsin2x/2-int(sin2xdx)/2=xsin2x/2+cos2x/4. Substitute for int(xcos2xdx): int(x^2sin2xdx)=-x^2cos2x/2+xsin2x/2+cos2x/4. Consider the limits -1 and 1. Rather than angles in degrees, it would seem more appropriate to consider angles in radians. Apply the limits: cos2=-0.416147, sin2=0.909297. (The expression becomes cos2/2+sin2/2+cos2/4-cos(-2)/2+sin(-2)/2-cos(-2)/4. Cosz=cos(-z) and sinz=-sin(z). The terms therefore cancel out.) The definite integral is 0.5587-0.5587=0, as expected.  
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Integration by Parts.

QUESTION: Integrate by parts the function 9x(ln(x))^2 wrt x between x = 1 and x = e^4. I = int 9x(ln(x))^2 dx let u = ln(x). When x = 1, u = 0, when x = e^4, u = 4 Then x = e^u and dx = e^u.du So now, I = int 9e^u.u^2.e^u du I = 9*int e^(2u).u^2 du integrating by parts, I/9 = (1/2)e^(2u).u^2 - int e^(2u).u du I/9 = (1/2)e^(2u).u^2 - {(1/2)e^(2u).u - int (1/2)e^(2u) du} I/9 = (1/2)e^(2u).u^2 - {(1/2)e^(2u).u - (1/4)e^(2u)} I/9 = (1/4)e^(2u){2u^2 - 2u + 1} I = (9/4)e^(2u){2u^2 - 2u + 1} Putting in the limits, u = 0 to u = 4 I = (9/4){[e^(0){0 - 0 + 1}] - [e^(8){32 - 8 + 1}]} I = (9/4){[1] - [e^8*25]} I = (9/4)(25.e^8 - 1)
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how do you integrate (x+1)cosx dx, using the symmerty rule and it has limits of (pi/2,-pi/2)

(x+1)cosxdx=xcosxdx+cosxdx. xcosxdx can be integrated by parts while cosxdx integrated is sinx. Let u=x and dv=cosxdx, so du=dx and v=sinx. d(uv)/dx=vdu/dx+udv/dx; integrating with respect to x: uv=int(sinxdx)+int(xcosxdx). So int(xcosxdx)=uv-int(sinxdx)=xsinx+cosx. Applying the limits: (pi)/2-(-(pi)/2*-1)=(pi)/2-(pi)/2=0.  xsinx+cosx is symmetrical. If f(x)=xsinx+cosx then f(-x)=(-x)(sin(-x))+cos(-x)=xsinx+cosx, because sin(-x)=-sin(x) and cos(-x)=cosx. Therefore f(x)=f(-x) hence f((pi)/2)=f(-(pi)/2) and f((pi)/2)-f(-(pi)/2)=0.
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calculate integration of log((2-x)/(2+x))

Question: calculate integration of log((2-x)/(2+x)) limits of integra1 to 1. ln[(2-x)/(2+x)] = ln(2-x) - ln(2+x) int ln[(2-x)/(2+x)] dx = int ln(2-x) - ln(2+x) dx = {-(2-x).ln(2-x) - x} - {(2+x).ln(2+x) - x} = -(2-x).ln(2-x) - (2+x).ln(2+x) Integrating between the limits of x = -1 to x = 1, I = {-(1).ln(1) - (3).ln(3)} - {-(3).ln(3) - (1).ln(1)} = {0 - ln(27)} - {-ln(27) - 0} = -ln(27) + ln(27) = 0 Answer: zero
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integration problems

Question: How can I write integration symbol? You can't! Not on this site I'm afraid. If you look on the menu bar above your text, when you are entering a problem, you will see an Omega symbol. (The Greek letter omega) Clicking on this symbol will open up a window from which you can choose special symbols. The integration symbol isn't among them. If you wish to enter an integration problem, in which, say, you wanted to integrate the function f(x), with respect to x, between the limits of a (lower limit) and b (upper limit), then you would enter it like this. int [ a to b] f(x) dx or, int [ a,b] f(x) dx HTH
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What is the capacity of the bowl?

A flat-bottomed circular bowl of parabolic cross-section has an interior base diameter of 16cm, an interior height of 10cm and its interior diameter across the top is 24cm. What is its capacity (volume) in litres? This bowl is a section of a solid parabola. Consider the volume of revolution of a parabola, y = a.x^2, about its vertical axis (y-axis), taken between the limits y1 and y2. We are given that the height of the bowl is h = y2 – y1 = 10 cm. Since the internal diameter, across the bottom of the bowl, is 16 cm, then the x-coordinate of this lower part of the bowl is x1 = 8 cm and its y-coordinate is y1. Since the internal diameter, across the top of the bowl, is 24 cm, then the x-coordinate of this upper part of the bowl is x2 = 12 cm and its y-coordinate is y2. We thus have two points on the bowl, P1(x1, y1) = (8, y1) and P2(x2, y2) = (12, y2) which are related by the equation, y=a.x^2. Using this equation with P1 and P2, y1 = a.x1^2 y2 = a.x2^2 Substituting values, y1 = a.64 y2 = a.144 Since the height of the bowl is given as 10 cm, then we have y2 – y1 = 10. Subtracting the previous two equations, y2 – y1 = 10 = 144a – 64a = 80a 10 = 80a a = 1/8 Then, y1 = 8 cm, y2 = 18 cm. Therefore the defining equation of the parabola is: y = (1/8)x^2. We can now re-describe the bowl as the volume of revolution of a parabola, y = (1/8)x^2, about its vertical axis (y-axis), taken between the limits y1 = 8 and y2 = 18. This volume is given by the formula, V = int [ y = y1 .. y2] (1/a).π.y dy. V = int [ y = 8 .. 18] (8).π.y dy. Integrating, V = [4.π.y^2] [18 .. 8] V = 4.π.{324 – 64} = 4.π.{260} V = 1040.π. cu cm. V = 3.267 litres
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integral e^(x^2+x)dx of x=0 to x=+infinity

$\int\limits_{0}^{+\infty }e^{x^{2}+x}dx$
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what is the limit of int x as x approaches 0 from the positive side?

"int" must be integer funkshun it meen round down 0.1 round down tu 0 0.9 round down tu 0 0.9999999 round down tu 0 thus, int(x) =0 as x get kloes tu zero from the rite from the left, int(x)=-1
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how do you intergrate (1-2cost)^2

QUESTION: how do you integrate (1-2cost)^2  with boundaries of 5pi/3 and pi/3. use this integral to find  the exact value of the shaded area. I = int (1 – cost)^2 dt I = int 1 – 2cost + cos^2t dt Using the identity cos2t = 2.cos^2t - 1 I = int 1 – 2cost + (1/2)(1 + cos2t) dt I = int 3/2 – 2cost + (1/2)cos2t dt I = 3t/2 – 2sint + (1/4)sin2t Using the limits t = pi/3 and t = 5.pi/3 I = {15.pi/6 – 2.sin(5.pi/3) + (1/4)sin(10.pi/3)} – {3.pi/6 – 2.sin(pi/3) + (1/4)sin(2.pi/3)} I = {15.pi/6 + sqrt(3) – sqrt(3)/8} – {3.pi/6 – sqrt(3) + sqrt(3)/8} I = 12.pi/6 +7.sqrt(3)/8 + 7.sqrt(3)/8 I = 2.pi + 7.sqrt(3)/4  
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limit from. 0 to infinity[ne^-x] where n is natural no.

?????? "natural number" ????????? wot lingo du yu yuze at skuel???? limit dont go from a tu b,...limit as yu gotu target, in this kase infinity limit as e^-x at x gotu infinity=0
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