2u questions (1 Viewer)

gummibear0731 · Dec 2, 2016

1. find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 to y=4

2a. differentiate (2x^2+1)/(3x^2-4)
b. hence evaluate (integral from 0 to 1) of x/(3x^2-4)^2 dx

P.s. for question 2 i can do part a but dont know how to do part b.

3. find the exact area bounded by the parabola y=x^2 and the line y=4-x

4. find the area enclosed between the curves y=(root)x and y=x^3

5. for the shaded region find the area and the volume around the x-axis. The region is of x^2=4ay coloured to (0,a)

These are from maths in focus. cheers for ANY help

android0 · Dec 3, 2016

1. You should draw a diagram. I would split the parabola into two functions by making x a function of y, where sqrt(y) is the principal root.

$x = +\sqrt y - 5$
$V_1$ = $\int^{4}_{1} \pi (\sqrt y - 5)^2, dy$

$x = -\sqrt y - 5$
$V_2$ = $\int^{4}_{1} \pi (-\sqrt y - 5)^2, dy$

Subtract V1 from V2.

2. $\frac{d(\frac{2x^2+1}{3x^2-4})}{dx}$ = $\frac{-22x}{(3x^2-4)^2}$

$\frac{2x^2+1}{(3x^2-4)}$ = $\int\frac{-22x}{(3x^2-4)^2}, dx$

$\int^{1}_{0}\frac{x}{(3x^2-4)^2}, dx $ = $[\frac{2x^2+1}{-22(3x^2-4)}]^1_0$

You can move constants outside of integrals.

3. To find where the two functions intersect: $$Set $x^2 = 4-x$

$Solve the quadratic for x, and your values become the bounds for the following integral:$

$\int(4-x)-x^2, dx$

4. $$Set $x^3 = \sqrt x$

$$Set $a = \sqrt x$

$a(a^5-1) = 0 $, a = 0, 1. $$

$\int^{1}_{0} \sqrt x - x^3,dx$

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2u questions (1 Viewer)

gummibear0731

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