integration probz (1 Viewer)

[sephiroth*]

Been bashing my head against a brick wall trying to answer these. helps

1) find the area bounded by the lines y=x + 1, x = -3, x=4
2) Find the area between the curves y=x^2 and y=x^4, between x= 1 and
x= 3 (how would u graph x^4?)
3)Find the area enclosed between the curve y=x^3-2 and the y-axis between y= -1 and y=25

Answers
1)14.5 units^2
2)39.7333 units^2
3)9.3333 units^2

Thanx

 

[DeathB4Life]

just started integration last week so i may not be 100%

y=x+1 is a diagnol straight line one unit "upwards"

integrated: (x^2)/2 + x

there are two sections to add up, one section below the x axis between -3 and -1 (x intercept) and the other above the x axis between -1 and +4

Total Area = |f(-3) - f (-1)| + |f(4) - f(-1)|

f(-3) = ((-3)^2)/2 + (-3)
= 1.5

f(-1)= ((-1)^2) + (-1)
= -0.5

f(4)= ((4)^2)/2 + 4
=12

|1.5 - (-0.5)| + |12 - (-0.5)|
= 2 + 12.5 = 14.5 units^2


y = x^4 should look like y = x^2 except alot more steeper.
ill attempt question 2, though we havent learnt it in class yet. question 3 i havent even got a clue whats being asked.

 

[michelle88]

hey, here's the answer to q2, i scanned it and you can view it from here:

http://i51.photobucket.com/albums/f384/michelledemler/scanintegrationprob.jpg

i got a different answer to q3 than you said, so i wont post my working out as i probably did something wrong, although i can't see where i did go wrong.

cya

 

[Mountain.Dew]

DeathB4Life, welcome to the forums!

2) Find the area between the curves y=x^2 and y=x^4, between x= 1 and
x= 3 (how would u graph x^4?)

y=x^4 has the same parabolic shape as y=x^2, except its a little bit steeper.

how to find the area is simple. realise that between x=1 and x=3, y=x^4 has higher y-values than x^2. So, the LENGTH of the 'strips' - that are dx thick - that we sum together, is (x^4 - x^2) long.

we sum all these 'strips' from x=1 to x=3. thus, applying the general formula that area=

b
∫ y dx , the area for this question is =
a

3
∫ (x^4 - x^2)dx = [x^5/5 - x^3/3]³₁
1

= [243/5 - 9] - [1/5 - 1/3] = 39.7333 units^2, as required.

 

[Mountain.Dew]

now, with q3, its a little bit harder.

after sketching the graph of x^3 - 2, we see that the curve cuts the x-axis at a certain pt, at x=2^1/3.

realise that if u evaluate the integral normally, like this:

25
∫ x^3 - 2 dx, it will obtain the wrong answer. this is because this integral will take one 'area' that is
-1

above the x-axis, off the other area below the x-axis.

so, to find the area, you need to evaluate TWO integrals.

 

[sephiroth*]

Thanks death, michelle n mountain. u took sum time to scan and do my probz , champion. cyas