- #1
Monoxdifly
MHB
- 288
- 0
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
My attempt so far:
$$x^2+2x-3=0$$
(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.
$$\int_1^2(x^2+2x-3)dx$$
$$=[\frac{1}{3}x^3+x^2-3x]_1^2$$
$$=(\frac{1}{3}(2^3)+2^2-3(2))-(\frac{1}{3}(1^3)+1^2-3(1)$$
$$=(\frac{8}{3}+4-6)-(\frac{1}{3}+2-3)$$
$$=(\frac{8}{3}+2)-(\frac{1}{3}-1)$$
$$=\frac{8}{3}+2-\frac{1}{3}+1$$
$$=\frac{7}{3}+3$$
$$=\frac{7}{3}+\frac{9}{3}$$
$$=\frac{16}{3}$$
$$=\frac{15}{3}$$
Where did I do wrong? I know from the graph that the answer is A since the area covered is less large than a $$1\times5$$ rectangle and the option A is the only one with a value less than 5, but what did I do wrong algebraically?
A. 4 area unit
B. 9 area unit
C. 11 area unit
D. 13 area unit
E. 27 area unit
My attempt so far:
$$x^2+2x-3=0$$
(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.
$$\int_1^2(x^2+2x-3)dx$$
$$=[\frac{1}{3}x^3+x^2-3x]_1^2$$
$$=(\frac{1}{3}(2^3)+2^2-3(2))-(\frac{1}{3}(1^3)+1^2-3(1)$$
$$=(\frac{8}{3}+4-6)-(\frac{1}{3}+2-3)$$
$$=(\frac{8}{3}+2)-(\frac{1}{3}-1)$$
$$=\frac{8}{3}+2-\frac{1}{3}+1$$
$$=\frac{7}{3}+3$$
$$=\frac{7}{3}+\frac{9}{3}$$
$$=\frac{16}{3}$$
$$=\frac{15}{3}$$
Where did I do wrong? I know from the graph that the answer is A since the area covered is less large than a $$1\times5$$ rectangle and the option A is the only one with a value less than 5, but what did I do wrong algebraically?