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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
$\displaystyle A = \int_0^2 |x^2+2x-3| \, dx = -\int_0^1 x^2+2x-3 \, dx + \int_1^2 x^2+2x-3 \, dx$
Attached Images area_parabola.jpg (18.9 KB, 0 views) 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. Tags integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jhenrique Calculus 5 June 30th, 2015 03:45 PM gen_shao Calculus 2 July 31st, 2013 09:54 PM maximus101 Calculus 0 March 4th, 2011 01:31 AM xsw001 Real Analysis 1 October 29th, 2010 07:27 PM maximus101 Algebra 0 December 31st, 1969 04:00 PM

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