Find the Area of x^2+2x-3 Region

[Monoxdifly]

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?

 

[skeeter]

... did you sketch a graph of the region in question? It exists between x = 0 (the y-axis) and x = 2

 

[Monoxdifly]

I did, I just forgot that the region outside the first quadrant also counted.