This is the entire question:

A curve has the equation y=(x+2)√(x-1).
i)Show that dy/dx = kx/√(x-1) and find k
ii)Hence, evaluate limit 2 to 5
∫ kx/√(x-1)dx

My answer: first part, i have already shown, and k is equals to 3/2

I don't get the second part though, it says evaluate ∫ kx/√(x-1)dx, isnt that equals to ∫ (dy/dx) dx, which is equals to y? o.o

correct. Evaluate y(5)-y(2)

To evaluate the integral ∫ kx/√(x-1)dx, we need to apply the fundamental theorem of calculus and use the result from part (i), which is dy/dx = kx/√(x-1).

The fundamental theorem of calculus states that if F(x) is an antiderivative of f(x) with respect to x, then ∫ f(x)dx from a to b is equal to F(b) - F(a). In other words, to evaluate the integral, we need to find an antiderivative of the integrand, which is kx/√(x-1).

In this case, we know that integrating kx/√(x-1) with respect to x gives us y, since dy/dx = kx/√(x-1). So we can rewrite the integral as ∫ dy, which will just give us y.

However, we need to evaluate the integral from 2 to 5, which means we need to find the value of y at those two points and then subtract them.

To find y, we can plug in the values of x into the equation y = (x+2)√(x-1):

For x = 2:
y = (2+2)√(2-1) = 4(1) = 4

For x = 5:
y = (5+2)√(5-1) = 7(2) = 14

Now, we can evaluate the integral:
∫ kx/√(x-1)dx from 2 to 5 is equal to the difference of y evaluated at 5 and y evaluated at 2:

∫ kx/√(x-1)dx from 2 to 5 = y(5) - y(2) = 14 - 4 = 10.

So the value of the integral is 10.