# Thread: An architect drawing !!!

1. ## An architect drawing !!!

i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

2. Originally Posted by carlasader
i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

The question as asked makes no sense. There must be at least one
constraint or additional conditions missing.

RonL

3. Originally Posted by carlasader
i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

Has this not already been posted?

RonL

4. 4. As $x^2 + x -1$ is positive between $x=1$ and $x=2$, the area is:

$
A = \int_1^2 (x^2 + x -1) dx
$

5. The first four terms in the expansion of $(1+x)(1-2x)^10$ may be found by expanding the second term using the binomial theorem and multiplying out:

$
(1+x)(1-2x)^{10}=(1+x)[1+10(-2x)+\frac{10.9}{2!}(-2x)^2+\frac{10.9.8}{3!}(-2x)^3+..]
$

and keeping the four terms with the lowest powers of $x$

RonL

5. ## thank u soo much for the answer

thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can

6. Originally Posted by carlasader
thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can
let's find the distance $PQ$ and $QM$ then, so we can find the area of the triangle.

to do that, we need to find points $P$ and $Q$

these points will be where the graphs intersect, that is, where

$x^2 - 2x + 2 = -x + 4$

$\Rightarrow x^2 - x - 2 = 0$

$\Rightarrow (x - 2)(x + 1) = 0$

$\Rightarrow x = 2 \mbox { or } x = -1$

when $x = 2$:

$y = -2 + 4 = 2$

when $x = -1$:

$y = 1 + 4 = 5$

so we have $P(-1,5)$ and $Q(2,2)$

Now, slope of line PQ is $\frac {y_2 - y_1}{x_2 - x_1} = \frac {3}{-3} = -1$

slope of line QM is $\frac {y_2 - y_1}{x_2 - x_1} = 1$

this means, line PQ is perpendicular to QM, and so $\triangle PQM$ is a right-triangle.

Now let's find the lengths of PQ and QM

By the distance formula:

$PQ = \sqrt {(2 - (-1))^2 + (2 - 5)^2} = \sqrt {18}$

$QM = \sqrt {(1 - 2)^2 + (1 - 2)^2} = \sqrt {2}$

$\Rightarrow A_{ \triangle PQM} = \frac {1}{2}bh = \frac {1}{2} \left( \sqrt {2} \right) \left( \sqrt {18} \right) = 3$

Now to find the area of the shaded region:

$A_{R} = \int_{-1}^{2} \left( -x + 4 - x^2 + 2x - 2 \right) dx$

$= \int_{-1}^{2} \left( -x^2 + x + 2 \right) dx$

$= \left[ - \frac {1}{3} x^3 + \frac {1}{2} x^2 + 2x \right]_{-1}^{2}$

$= 4 \frac {1}{2}$

Since $4 \frac {1}{2} = 1 \frac {1}{2} \cdot 3$, we have the desired result