Mathematics

# If $\displaystyle P = \int_0^2 \frac{17+8x-4x^2}{e^{6(1-x)}+1}dx$, then find the value of $\left[\displaystyle \frac{21P}{118}\right]$, where [ ] denotes greatest integer function.

3

##### SOLUTION
Let $P=\displaystyle\int_{0}^{2} \dfrac{17+8x-4x^2}{e^{6(1-x)} +1} dx=P_1(say)$

Using, $\int_{a}^{b} f(x)dx=\int_{a}^{b} f(a+b-x)dx$

$P=\displaystyle\int_{0}^{2} \dfrac{17+8(2-x)-4(2-x)^2}{e^{6(1-(2-x))} +1} dx$

$P=\displaystyle\int_{0}^{2} \dfrac{17+8x-4x^2}{\dfrac{1}{e^{6(1-x)}} +1} dx$

$P=\displaystyle\int_{0}^{2} e^{6(1-x)}. \dfrac{17+8x-4x^2}{e^{6(1-x)} +1} dx=P_2(say)$

$2P=P_1+P_2=\displaystyle\int_{0}^{2} \left(e^{6(1-x)}+1\right). \dfrac{17+8x-4x^2}{e^{6(1-x)}+1}dx$

$\implies2P=\displaystyle\int_{0}^{2} (17+8x-4x^2)dx$

$\implies2P=17[x]_{0}^{2}+4[x^2]_{0}^{2}-\dfrac{4}{3} [x^3]_{0}^{2}$

$\implies 2P=17\times 2+4\times 4-\dfrac{4}{3}\times8$

$\implies2P=\dfrac{118}{3}\implies P=\dfrac{59}{3}$

$\implies \dfrac{21P}{118}=\dfrac{21}{118}\times\dfrac{59}{3}=3.5$

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

Hence, the correct answer is $3$.

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One Word Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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