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March 18th, 2018, 08:34 AM   #1
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[ASK] Integral

The area of the region limited by the curve $\displaystyle y=x^2+2x-3$, X-axis, Y-axis, and the line x = 2 is ....
A. 4 area unit
B. 9 area unit
C. 11 area unit
D. 13 area unit
E. 27 area unit

My attempt so far:
$\displaystyle x^2+2x-3=0$
(x + 3)(x - 1) = 0
x = -3 or x = 1
X-intercept is at x = -3 and x = 1.
After drawing the graph, the restriction is x = 1 to x = 2.
$\displaystyle \int_1^2(x^2+2x-3)dx$
$\displaystyle =[\frac{1}{3}x^3+x^2-3x]_1^2$
$\displaystyle =(\frac{1}{3}(2^3)+2^2-3(2))-(\frac{1}{3}(1^3)+1^2-3(1)$
$\displaystyle =(\frac{8}{3}+4-6)-(\frac{1}{3}+2-3)$
$\displaystyle =(\frac{8}{3}+2)-(\frac{1}{3}-1)$
$\displaystyle =\frac{8}{3}+2-\frac{1}{3}+1$
$\displaystyle =\frac{7}{3}+3$
$\displaystyle =\frac{7}{3}+\frac{9}{3}$
$\displaystyle =\frac{16}{3}$
$\displaystyle =\frac{15}{3}$
Where did I do wrong? I know from the graph that the answer is A since the area covered is less large than a $\displaystyle 1\times5$ rectangle and the option A is the only one with a value less than 5, but what did I do wrong algebraically?
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March 18th, 2018, 09:24 AM   #2
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Quote:
The area of the region limited by the curve $y=x^2+2x−3$, X-axis, Y-axis, and the line x = 2 is ....
A. 4 area unit
B. 9 area unit
C. 11 area unit
D. 13 area unit
E. 27 area unit
$\displaystyle A = \int_0^2 |x^2+2x-3| \, dx = -\int_0^1 x^2+2x-3 \, dx + \int_1^2 x^2+2x-3 \, dx$
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March 18th, 2018, 03:50 PM   #3
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Damn, I forgot that it doesn't have to be in the first quadrant. Thanks skeeter.
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