- #1

Monoxdifly

MHB

- 284

- 0

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?