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 $\displaystyle y=x^2+2x3$, Xaxis, Yaxis, 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: $\displaystyle x^2+2x3=0$ (x + 3)(x  1) = 0 x = 3 or x = 1 Xintercept is at x = 3 and x = 1. After drawing the graph, the restriction is x = 1 to x = 2. $\displaystyle \int_1^2(x^2+2x3)dx$ $\displaystyle =[\frac{1}{3}x^3+x^23x]_1^2$ $\displaystyle =(\frac{1}{3}(2^3)+2^23(2))(\frac{1}{3}(1^3)+1^23(1)$ $\displaystyle =(\frac{8}{3}+46)(\frac{1}{3}+23)$ $\displaystyle =(\frac{8}{3}+2)(\frac{1}{3}1)$ $\displaystyle =\frac{8}{3}+2\frac{1}{3}+1$ $\displaystyle =\frac{7}{3}+3$ $\displaystyle =\frac{7}{3}+\frac{9}{3}$ $\displaystyle =\frac{16}{3}$ $\displaystyle =\frac{15}{3}$ 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  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588  Quote:
 
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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