Question

Find the area bounded by the curves y=4/x^2, x-2y+7 and x=4.

Answer 1

At the intersection of the curves  

$y=\frac4{x^2}$ and $x-2y+7=0$ we have $\frac 4{x^2} = \frac {x+7}2$

0r $8=x^3+7x$ or $x^3+7x-8=0 …(i)$

Which is a cubic equitation with x=1 as a root.

We thus use synthetic division to find other roots (if any).

$\begin {array}{c| rrr}&1&1&0&-8\\-&&1&1&8\\\hline\\&1&1&2&0\\\end{array}$

$⇒x^3+7x-8= (x-1)(x^2+x-8)=0$

The function   $x^2+x+8$ has no real roots suggesting that the curves  $y= \frac 4{x^2}$ and

$x+2y+7=0$ intersect at a point where x=1.

$⇒area= ∫_1^4(\frac {x+7}2 - \frac 4{x^2})dx$

$=[\frac{ x^2}{4}+ \frac 72 x+\frac 4x]_1^{4}$

$=9-\frac {31}4$

$=11.25 \, square\, units\, of \, length$