# Thread: integrals and area

1. ## integrals and area

need help with these pls

integrate bewtween y=|x^2 -3x +2| and the x axis, between x=0 and x=3
area between x^2=4y and y^2=4x (confused about if we should integrate y or x)
area between y=-x^2+4x-3, y axis and the lines x=4 and y=4

2. ## Re: integrals and area

1. The parabola $x^2 - 3x + 2$ is negative when $1 < x < 2$. So when you take the absolute value, the only change in the graph is that that area will be flipped across the x axis to become positive. Same area, just the direction is changed. Therefore, to obtain the area between $| x^2 - 3x + 2|$ and the x-axis from $x=0$ to $3$, we can integrate three parts of the curve separately (not to be confused with integration by parts...):

$\int_0^1 x^2 - 3x + 2 \, dx + \left|\int_1^2 x^2 - 3x + 2 \, dx\right| + \int_2^3 x^2 - 3x + 2 \, dx$

Normally the middle integral would be negative, since the curve is under the x-axis at that point, but we can just take the absolute value.
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2. It may be less confusing to integrate on the x-axis, but it doesn't matter which you pick, you just have to rearrange both equations in terms of the same variable.

The parabola $y^2 = 4x$ is just the parabola $x^2 = 4y$ rotated 90 degrees clockwise about the origin.

To obtain the area where they intersect, we can rearrange $y^2 = 4x$ to make $y$ the subject, giving $y = \pm 2\sqrt{x}$. Since these parabolas never intersect below the x-axis, we actually don't need the $\pm$ sign so we can remove it and take the positive root.

Clearly, these integrals intersect at the origin and one other point, which is $(4,4)$. Therefore (keeping in mind which curve is above the other) we can use the integral

$\int_0^4 (2\sqrt{x}) - \left(\frac{1}{4}x^2\right) \, dx$
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3. If you draw a quick sketch it's easy to see that you are essentially being asked to find the area between $y=4$ and $y=-x^2+4x-3$ from $x= 0$ to $4$.

The integral we need is just

$\int_0^4 4 - (-x^2+4x-3) \, dx$

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