Integration-area question (1 Viewer)

[Loz_metalhead]

We have just started the integration topic. I could do the other area questions, but I think because this one has a 0 you must do something else.

Find the area bounded by the graph of each of the following functions, the x-axis and the given ordinates.

y=x^2-4; x=0 to x=3

and this one:
f(x)=2-x-x^2;x=0 to x=2

Thanks alot

 

[Riviet]

For the first, do the integral as normal, since the area between x=0 and x=3 turns out be all below the x-axis; your answer will be negative but it's an area so you need to take the absolute value this for your final answer.

For the second, the concave down parabola intercepts the x-axis at two points, the one we're worried about is the x-intercept at (1,0). Since the area between x=0 and 1 is above the x-axis and the area between x=1 and 2 is below it, you need to split the integral.

So find the integral from 0 to 1 of f(x) dx as normal and add it to |the integral from 1 to 2 of f(x) dx|. The absolute value is taken because the second area is below the x-axis as I mentioned before.
Hope that helps. =D

Tip: it never hurts to draw a diagram (ie the curve).

 

[Loz_metalhead]

Thanks. For the first one you actually dont treat it as a normal integral, you do it the same as the second.

the answers are 7 2/3 and 3

 

[SoulSearcher]

the answer at the back of the book is right

you see, the curve y = x² - 4 cuts the x-axis at 2

since the area you're looking for is between x = 0 and x = 3, then you would have to find the absolute value of the area between x = 0 and x = 2 and add that value to the value of tha area between x = 2 and x = 3

 

[Loz_metalhead]

Yep, thanks. It really helps when you draw a diagram and shade it in.

 

[SoulSearcher]

Yep, thanks. It really helps when you draw a diagram and shade it in.

That always helps, even in exams with time limits

 

[pLuvia]

You should always draw out the graphs, you never know when the area is a negative which can alter your answer completely