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.


ANSWER

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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