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 March 18th, 2018, 08:34 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [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?
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 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Damn, I forgot that it doesn't have to be in the first quadrant. Thanks skeeter.

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