Question

If $P={\int }_0^2\frac{17+8x-4x^2}{e^{6(1-x)}+1}dx$, then find the value of [$\frac{21P}{118}$],

• A. $3.5$
• B. $3$
• C. $3.2$
• D. $2.8$

Answer 1

Answer: (A)

$3.5$

$P={\int }_0^2\frac{17+8x-4x^2}{e^{6(1-x)}+1}dx$
$P={\int }_0^2\frac{17+8(2-x)-4(2-x{)}^2}{e^{6(1-2+x)}+1}dx$
(By property)
Add $2P={\int }_0^2(17+8x-4x^2)dx$
$\Rightarrow P=\frac{1}{2}(17x+4x^2-\frac{4x^3}{3}{)}_0^2$
$\Rightarrow P=\frac{59}{3}\therefore \left[\frac{21P}{118}\right]=3.5$