a) find the number (A) such that the line x= (A) bisects the area under the curve y= 1/x^2, 1 less than or equal to X which is less than or equal to 4
b)find the number (B) such that the line y=(B) bisects the area in part a)
a) find the number (A) such that the line x= (A) bisects the area under the curve y= 1/x^2, 1 less than or equal to X which is less than or equal to 4
b)find the number (B) such that the line y=(B) bisects the area in part a)
Determine area:
a...
$\displaystyle \int_1^4 \frac{1}{x^2}dx=\frac{3}{4}$
We want the area that bisects ie. half of the area found above
$\displaystyle \int_1^t\frac{1}{x^2}dx=\frac{3}{8} \Rightarrow t=\frac{8}{5}$
So $\displaystyle x=\frac{8}{5}$
b...
again we want to bisect:
$\displaystyle \int_1^4\frac{1}{x^2}-b \quad dx=\frac{3}{8} \Rightarrow b=\frac{1}{8}$