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

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In summary, the area of the region limited by the curve y=x^2+2x-3, X-axis, Y-axis, and the line x = 2 is 4 units. After drawing the graph, it is clear that the region also exists outside of the first quadrant, which was initially overlooked. Using the integral method, the correct answer is obtained by finding the area between x = 0 and x = 2, which is 4 units.
  • #1
Monoxdifly
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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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  • #2
... did you sketch a graph of the region in question? It exists between x = 0 (the y-axis) and x = 2
 
  • #3
I did, I just forgot that the region outside the first quadrant also counted.
 

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

1. What is the formula for finding the area of x^2+2x-3 region?

The formula for finding the area of x^2+2x-3 region is to first factor the equation into (x+3)(x-1) and then use the formula for finding the area of a rectangle, which is length times width. In this case, the length would be (x+3) and the width would be (x-1).

2. How do you determine the boundaries for finding the area of x^2+2x-3 region?

The boundaries for finding the area of x^2+2x-3 region can be determined by finding the x-intercepts of the equation, which are the points where the graph crosses the x-axis. In this case, the x-intercepts are at x=-3 and x=1.

3. Can the area of x^2+2x-3 region be negative?

No, the area of x^2+2x-3 region cannot be negative. The area is always a positive value, representing the amount of space within the boundaries of the region.

4. Is it necessary to use calculus to find the area of x^2+2x-3 region?

No, it is not necessary to use calculus to find the area of x^2+2x-3 region. As mentioned before, the formula for finding the area of a rectangle can be used to find the area of this region without the use of calculus.

5. Can the area of x^2+2x-3 region be infinite?

No, the area of x^2+2x-3 region cannot be infinite. The region is bounded by the x-axis and the two vertical lines at x=-3 and x=1, so the area is finite.

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