Integration of a curve in the 2nd quadrant. (1 Viewer)

summonbolt · Feb 1, 2017

Hi! I was just wondering how to do the following question:

Find the area enclosed between the lines y=4 and y=1-x in the second quadrant
using integration. My teacher solved the question by doing the area as a triangle, but I was just wondering if there's a way to solve using integration. Thanks!

InteGrand · Feb 1, 2017

summonbolt said:
Hi! I was just wondering how to do the following question:

Find the area enclosed between the lines y=4 and y=1-x in the second quadrant
using integration. My teacher solved the question by doing the area as a triangle, but I was just wondering if there's a way to solve using integration. Thanks!

$\noindent The lines intersect at $x = -3$, so we are after the area between the curve $y=1-x$ and $y=4$ from $x=-3$ to $0$. The upper curve is always $y=4$ here and the lower curve $y= 1-x$, and so the area could be computed as $\int_{-3}^{0} \left(4-(1-x)\right)\,\mathrm{d}x$.$

RickB · Feb 1, 2017

Hi,

The graphs of y = 4 and y = 1 - x meet at (-3, 4). The required area will be (area from y = 4 down to the x-axis) MINUS (area from y = 1 - x down to x-axis), starting at x = -3 and ending at x = 0. So, your calculation will be

INTEGRAL (-3 to 0) of 4 MINUS INTEGRAL (-3 to 0) of 1 - x.

Should give 4.5 which is the area.

Integration of a curve in the 2nd quadrant. (1 Viewer)

summonbolt

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InteGrand

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RickB

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