integration baby (1 Viewer)

darshil · May 18, 2009

Hiya everyone,
i just cannot algebraically solve these:

1) calculate the area of the region enclosed by he line y=x+1 and the parabola y=((x^2)/4)-2

(the brackets are not actually there in the question)

2) Calculate the area bounded by the grapshs of f(x)=x^2, g(x)=1/(x^2), x>0 and the line x=3

Thanks a lot everyone, will never forget the help

kurt.physics · May 18, 2009

darshil said:
Hiya everyone,
i just cannot algebraically solve these:

1) calculate the area of the region enclosed by he line y=x+1 and the parabola y=((x^2)/4)-2

(the brackets are not actually there in the question)

Thanks a lot everyone, will never forget the help

$Solve these two graphs simultaneously to determine where they intersect$

$x+1 = \frac{x^2}{4} - 2$

$\frac{x^2}{4} -x -3 = 0$

$x^2 -4x -12 = 0$

$(x-6)(x+2) = 0$

$\therefore x = 6,-2$

Between -2 and 6, the line is greater than the quadratic

<x><x> $\therefore A = \int_{-2}^{6} \left( (x+1) - (\frac{x^2}{4}-2) \right) dx$

</x></x> $= \left[ \frac{x^2}{2} + x - \frac{x^3}{12} + 2x \right]_{-2}^{6}$

$= \left[ \frac{x^2}{2} - \frac{x^3}{12} +3x \right]_{-2}^{6}$

$= 21 \frac{1}{3}$

kurt.physics · May 18, 2009

darshil said:
Hiya everyone,
i just cannot algebraically solve these:

2) Calculate the area bounded by the graphs of f(x)=x^2, g(x)=1/(x^2), x>0 and the line x=3

Thanks a lot everyone, will never forget the help

$You have to find where f(x) and g(x) intersect, drawing the graphs would also help$

$You would find that the only area that is bounded by the three curves is between 1 and 3$

$Between 1 and 3, f(x) is greater than g(x)$

$\therefore A = \int_{1}^{3} \left({x^2 - \frac{1}{x^2} \right) dx$

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

$=8$

darshil · May 19, 2009

Thanks for the replies man, I really appreciate it !

PS: How do you know the line is greater than the quadratic?

kurt.physics · May 19, 2009

darshil said:
Thanks for the replies man, I really appreciate it !

PS: How do you know the line is greater than the quadratic?

When you draw the graph, it becomes immediately obvious. But the line y = x+1 is only greater than y= (x^2/4)-2 between -2 and 6, for the rest, the quadratic is greater than y= x+1.

darshil · May 20, 2009

Gotchya man ! Thanks yet again

Just one thing, you forgot to add the units squared in the answers (= (nonetheless the right answer)

Timothy.Siu · May 20, 2009

darshil said:
Gotchya man ! Thanks yet again

Just one thing, you forgot to add the units squared in the answers (= (nonetheless the right answer)

our teacher tells us not to lol

Gibbatron · May 20, 2009

I think you can actually lose marks in the exam if you dont put the units there but, so might have to be careful there.

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