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Integrate y=x^2 between the limits x=3 and x=1

Integrate y=x^2 which is 1/3 x^3 Subsitute the limits, (1/3 (3)^3)-(1/3 (1)^3) 27/3 - 1/3 = 26/3
CW
Answered by Caleb W. • Maths tutor
4234 Views

Given that y=((3x+1)^2)*cos(3x), find dy/dx.

As why is in the for y=uv where u and v are funtions of x, dy/dx=u'v+v'u (where ' implies the derivative) u=(3x+1) 2 , v=cos(3x) therefore using the chain rule u'=2 3 (3x+1)=18x+6 and v'=-3sin(3x). Using thi...
WR
Answered by William R. • Maths tutor
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Integral of e^x*sinx

written out
JJ
Answered by Jamil J. • Maths tutor
4425 Views

integrate with respect to x the function f(x)= xln(x)

Use integration by parts let u=ln(x) let dv/dx=x therefore du/dx=1/x and v=(1/2)x^2 therefore the integral of xln(x) is equal to the following: (1/2)x^2ln(x) - (integral with respect to x of:((1/2)x^2)/x) = ...
PJ
Answered by Priya J. • Maths tutor
3515 Views

Differentiate: (12x^3)+ 4x + 7

36x^2 + 4
SP
Answered by Saskia P. • Maths tutor
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