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split (x^2 – 2x +3) / x^4 into: (x^-2 -2x^-3 + 3x^-4) dx = -x^-1 + x^-2 – x^-3 + c = -1/x + 1/x^2 – 1/x^3 + c

at exact x=a the function is not defined. but the left hand limit (LHL) at x=a gives a very small negative number and RHL give a very small positive number. Hence the function is discountinuous at...

xy-3x-2y-10=0Take y common,y(x – 2) = 3x + 10y = (3x + 10)/(x-2)Point of discontinuity is at x = 2It represents a curve of which relation. in y and x with other than the ellipse, parabola, hyperbola...

Do it by partial fraction, ∫(1- x 2 /x-2x 2 )dx can be written as: (1-x)*(1 +x)/x*(1 -2x)dx = A/x + B/(1 -2x) (1 -x) (1 +x) =A(1- 2x) + B(x) Put x= 1, -1 and get the values of A nd B and then...

The above integeral can be solved by using Partial Fraction ------------> ∫ (1dx)/(a-x)(b-x) = A/(a -x) + B/(b -x) 1 = A(b -x) + B(a – x) Now find the values of A and B. 1. put x=a 1 = A(b – a) A =...

U will havee to check first if the given function to itegerate is even or odd? If it is odd then the integeral becomes zero automatically. If it is even then, he integeral becomes: 2*f(x)dx ------>...

Apply By part integeration ----> Let the first function be unity and second one root sinx. integration of square root of sin x dx integration (sin x)^1/2 dx = [{(sin x)^3/2) / (3/2)}* cos x = [2cos...

area is never negative . so area should always be calc with modulus function . while integrating , we must split the integral at the points where the curves intersect . hope this helps you.

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integration is the act of integrating,it is the act of bringing together smaller components into a single or small type of functions as only one..........

this is based on knowledge of differntiaton.as 1/x is of the form of x**-1.differntiation in a nutshell is the opposite of integration. so as d/dx of ln(x) is 1/x.hence integral of 1/x is ln(x). idk...

Intergration is also called anti-derivative so its just reverse of differentiation Thanks

go to bed at time and donot study till late night health is wealth so take a balanced diet no new concepts at this time just proper sleep and fresh mind helps you to give your best best of luck...

dear student, we know this formula ∫uv' dx = uv − ∫u'v dx now put v’= w So , ∫uw dx = uv − ∫u’ (∫w dx) dx Now replace w=v So, ∫uv dx = u∫v...

very lengthy question... assuming you are not a mathematitian and are studying for marks, here is a tip, if you see this type of long question, quickly skip it, it doesnt matter if you can do...

Dear student Your question is not complete.Please type the correct question in order to get a to-the-point answer.

hello anil kumar, where is the attachment quetion please attach the quetion..............................................................................................................

hello anil kumar, please tell me the quetion first.and type clearly the quetion..............................................................................

Ok now lets proceed step by step Variables: = R = radius of the semicircle, length = l and height = h Assume semicircle on the length => 2R = l and the perimeter is known hence P = constant = Now...

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