Question

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

Answer 1

Let $P={\int }_0^2\frac{17+8x-4x^2}{e^{6(1-x)}+1}dx=P_1\left(say\right)$

Using, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$P={\int }_0^2\frac{17+8(2-x)-4(2-x{)}^2}{e^{6(1-(2-x\left)\right)}+1}dx$

$P={\int }_0^2\frac{17+8x-4x^2}{\frac{1}{e^{6(1-x)}}+1}dx$

$P={\int }_0^2{e}^{6(1-x)}.\frac{17+8x-4x^2}{e^{6(1-x)}+1}dx=P_2\left(say\right)$

$2P=P_1+P_2={\int }_0^2(e^{6(1-x)}+1).\frac{17+8x-4x^2}{e^{6(1-x)}+1}dx$

$⟹2P={\int }_0^2(17+8x-4x^2)dx$

$⟹2P=17[x{]}_0^2+4[x^2{]}_0^2-\frac{4}{3}[x^3{]}_0^2$

$⟹2P=17\times 2+4\times 4-\frac{4}{3}\times 8$

$⟹2P=\frac{118}{3}⟹P=\frac{59}{3}$

$⟹\frac{21P}{118}=\frac{21}{118}\times \frac{59}{3}=3.5$

Hence, $\left[\frac{21P}{118}\right]=3$

Hence, the correct answer is $3$.