Q117 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

$$ \int_0^2 x^3(2-x)^4 d x= $$

  • A
    $\frac{128}{105}$
  • B
    $\frac{16}{35}$
  • C
    $\frac{256}{105}$
  • Correct answer
    D
    $\frac{32}{35}$

Solution

$\begin{aligned} & \text {} I=\int_0^2 x^3(2-x)^4 d x \Rightarrow I=\int_0^2(2-x)^3 \cdot x^4 d x \\ & \Rightarrow \quad 2 I=\int_0^2\left(x^3(2-x)^4+(2-x)^3 x^4\right) d x \\ & \Rightarrow \quad I I=\int_0^2 x^3(2-x)^3(2-x+x) d x \\ & \Rightarrow \quad I=\int_0^2 x^3(2-x)^3 d x \\ & \Rightarrow I=\int_0^2 x^3\left(8-12 x+6 x^2-x^3\right) d x \\ & \Rightarrow I=\int_0^2\left(8 x^3-12 x^4+6 x^5-x^6\right) d x \\ & \Rightarrow I=\left[2 x^4-\frac{12 x^5}{5}+x^6-\frac{x^7}{7}\right]_0^2 \\ & \Rightarrow I=32-\frac{384}{5}+64-\frac{128}{7}=\frac{32}{35} .\end{aligned}$

Hence, the correct answer is option: 'D'.

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