Math Help Needed, Any Help Appreciated?

Question

how do I figure out....

the square root of [x^2 - 4x + 4] dx over the intergal of 0 and 2

and...

If over the intergal of 2 and 4 f(x) dx = 6, then over the intergal of 2 and 4 (f(x)+3) dx =?

any help will help thanks

Answer 1

I don't understand the whole question.

square root of (x^2 -4x +4) is (x-2)

normally, the integral is taken of a function, multiplied by dx.

You can integrate over a finite distance, for example, from 0 to 2 (meaning that you will evaluate the resulting Function (after integrating the original function) by comparing its value when x=2 and its value when x=0.

If what you seek is:
integral from 0 to 2 of (x-2)dx

then:
integral of (x-2)dx = x^2 / 2 - 2x +C
where C is a coefficient.
Normally, for definite integrals, we don't bother with the coefficient. This time I'll leave it in so that you will see why it disappears.

Evaluate the resulting Function at the 'end' value of the integral x=2:
x^2 / 2 - 2x +C = 4/2 - 4 + C = -2 +C

Evaluate the Function at the 'start' value x = 0

x^2 / 2 - 2x +C = 0 + C

Subtract the 'start' from the 'end':
-2 + C - (0 + C) = -2 + C - C = -2

(The C always cancels out)

-----

The second one appears to be:

If the integral from 2 to 4 of f(x)dx is equal to 6, what is the integral, also from 2 to 4, of [f(x) + 3]dx.

Here, the trick is to know that you can 'distribute' the operator:

integral of [f(x) + 3]dx = integral of f(x) dx + integral of 3 dx

integral from 2 to 4 of [f(x) + 3]dx =
integral from 2 to 4 of f(x) dx (which we know is 6) +
integral from 2 to 4 of 3 dx.

integral of 3 dx = 3x (+ C, which we forget)

3x evaluated at x=4 is 12
3x evaluated at x=2 is 6
The difference is 6
(higher value of x minus lower value of x -- be careful because sometimes the higher value of x gives a lower value for the Function; then you might get a negative difference)

Therefore:
integral from 0 to 2 of [f(x) + 3]dx = 6 + 6

Answer 2

2
∫(x^2 - 4x + 4)^1/2 dx = 4
0

(x^2 - 4x + 4)^1/2 makes a triangle with vertices (0,2), (0,-2), and 2,0 The area of this triangle is 2*4/2 = 4
2
∫(x - 2)dx = (1/2)x^2 - 2x |(0,2) = 2 - 4 = -2, or 2
0
square units below the x-axis

2
∫(-x + 2)dx = (-1/2)x^2 + 2x |(0,2) = 2 - 4 = 2 square
0
units above the x-axis. The area bounded by the curve and the y-axis is then 2 + 2 = 4 square units.

and...

given
4
∫f(x)dx = 6
2

4
∫(f(x) + 3)dx = 6 + 12 - 6 = 12
2

Answer 3

For the first problem

x^2-4x+4 factors as (x-2)^2, then the integral simplifies to

integral from 0 to 2 of square root of ((x-2)^2) and this is the
integral from 0 to 2 of x-2 =
x^2-2x evaluated from 0 to 2 which works out to be 0.


For the second problem
The integral from 2 to 4 of (f(x)+3) dx is the same as

The integral from 2 to 4 of f(x) + the integral from 2 to 4 of 3 dx

We know the first part is 6 so now you find the integral of 3 from 2 to 4, which is 3x evaluated from 2 to 4 which is 6.

So overall the integral from 2 to 4 of (f(x)+3) =6+6 =12

Answer 4

Consider this:

sqrt(x^2 - 4x + 4) = sqrt(u) = u^1/2, u = x^2 -4x + 4

Int[0-2] u^1/2 du = 2/3 u^3/2 |2|0

2/3(2^2 - 4(2) + 4) - 2/3(0^2 - 4(0) + 4)

2/3(0) - 2/3(4) = 8/3

And then:

Int[2-4](f(x) + 3)dx, Int[2-4](f(x))dx = 6

Int[2-4](f(x))dx + 3Int[2-4]dx

6 + (3/2(x^2)|4|2)

6 + (3/2(4^2) - 3/2(2^2))

6 + (24 - 6)

6 + 18 = 24

Answer 5

sqrt(x^2 - 4x + 4)=|x-2|

integral of |x-2| over 0 and 2
-x+2 over 0 and 2

gives 2

f(x) dx = 6
f(x)=|x-2|
(4-2)*y/2=6
f(4)=6

integral (|x-2|+3) dx over 2 and 4 is
=(4+1)-(2+1)=2

Answer 6

the square root of (x^2-4x+4) is (x-2).
The integral of (x-2)dx is (x^2)/2-2x+c
The integral from 0 to 2 is -2


0

Answer 7

you should be looking on your math e book. you in ordinary terms learn from expereince. yet im in a sturdy temper: you upload up 40+40 two=80 two one hundred eighty-80 two=ninety 8 so the three angles are 40 40 two & ninety 8 because of the fact that your working with angles its the two going to be an acute, suitable, or obtuse triangle. this triangle could be an obtuse triangle because of the fact a minimum of one in each of its angles is obtuse (ninety 8). decrease than ninety is acute; above ninety is obtuse.

Answer 8

i think you should do it by your self keep on trying
SORRY