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Find the area under the curve of y=1/(3x-2)^0.5 between the limits x=1 and x=2 and the line y=0

This question requires integration since the area under the curve is equal to the integral between these bounds. Initially let u=3x-2 and differentiate with respect to x so then du/dx = 3. Rearrange to dx =d...
CT
Answered by Callum T. • Maths tutor
3775 Views

At time t = 0 a particle leaves the origin and moves along the x-axis. At time t seconds, the velocity of P is v m/s in the positive x direction, where v=4t^2–13t+2. How far does it travel between the times t1 and t2 at which it is at rest?

First, you have to work out the values of t1 and t2 at which the particle is at rest. This is done by solving the quadratic equation for v, producing values for t of 13/8 +- sqrt(137): 0.1619s and 3.088s. To...
JM
Answered by Jack M. • Maths tutor
6908 Views

Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)

We must first complete the square for both equations and get the equations in the (x-a) 2 +(y-b) 2 =r 2 with the primary objective of determining the centres of both circles (a,b) . Through completing the sq...
SS
Answered by Shanti S. • Maths tutor
4160 Views

How do you differentiate y=ln(x)

I would use the fact that ln is the inverse function of the exponential function e^x to re-write the equation as x=e^y. This can be directly seen by just putting e^y=e^(lnx). Since the definition of a ln(a)=...
MG
Answered by Milo G. • Maths tutor
13854 Views

Find the area bounded by the curve y=(sin(x))^2 and the x-axis, between the points x=0 and x=pi/2

First, use the identity cos(2x)=(cos(x))^2-(sin(x))^2 along with the identity (sin(x))^2+(cos(x))^2=1 to obtain the integral of 1/2*(1-cos(2x)) as it is not possible to integrate (sin(x))^2 straight off with...
TL
Answered by Thomas L. • Maths tutor
5552 Views