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                   area under curve
                   

3 years ago

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Are of the cure is the integral of the graph function

3 years ago

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∫ -a a e ax cosnxdx
 
 
see , you have to use by parts.. .. for simplification put “ nx “ = “t” … then dx = dt/n , and e ax = e (a/n)*t now , Take COSt as the first part and e (a/n)*t as the second part .. I =∫ -a...
 
abhidon3 4 months ago
evaluate ∫ (sqrt(x)( ax^2 + bx + c)) dx
 
 
∫ (sqrt(x)( ax^2 + bx + c)dx =​∫ (x^1/2(ax^2 +bx +c)dx =​∫ (ax^5/2 +bx^3/2 +cx^1/2)dx =2​ax^7/2 /7 +2bx^5/2 /5 +2cx^3/2 /3 +K {we know that​∫x^ndx =(x^(n+1) /(n+1)} Here K is...
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Sunil Raikwar 6 months ago
 
make ax 2 +bx+c in to form of a((x-b/2) 2 +(c -b 2 /4)). then put (x-b/2)=(c-b 2 /4)^.5*tany. then integrate it. simple
 
Karthick vel 6 months ago
Integrate from 0 to infinity ∫[3/(x 2 +1)] dx [] denotes the greatest integer function..
 
 
Ans: Hello Student, Please find answer to your question below Here,
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Jitender Singh 7 months ago
 
Thank u Jitender Sir.
 
Lovey 7 months ago
Value of x in the following –
 
 
Image is attached
 
Drake 8 months ago
Q. THE RANGE OF THE FUNCTION sin(inverse)( x(square) + 2x)
 
 
Ans: Hello Student, Please find answer to your question below First we need to check the domain of the function ….......(1) So the domain of f(x) is (1). Since it covers all the...
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Jitender Singh 6 months ago
FIND UNIT VECTORS PERPENDICULAR TO THE PLANE DETERMINED BY THE POINTS P(1,-1,2) , Q(2,0,-1) AND R(0,2,1).
 
 
The two vectors in the given plane is PQ & PR PQ = (2 – 1) i + (0 + 1) j + (-1 – 2)k = i + j – 3k PR = (0 – 1)i + (2 + 1) j + (1 – 2k = –i + 3j – k The vector perpendicular to PQ & PR is PQ...
 
Y RAJYALAKSHMI 7 months ago
 
THANKS!!!!
 
bharat makkar 7 months ago
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