Q111 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

If $\int \phi(x) d x=\psi(x)$, then $\int\left(\phi_0 h\right)(x) h(x) h^{\prime}(x) d x=$

  • A
    $(\phi h)(x) \phi^{\prime}(x)-\int(\phi h)(x) h^{\prime}(x) d x+c$
  • Correct answer
    B
    $\left(\psi_0 h\right)(x) h(x)-\int\left(\psi_0 h\right)(x) h^{\prime}(x) d x+c$
  • C
    $\left(\psi_0 h\right)(x) \phi(x)-\int\left(\psi_0 h\right)(x) \phi^{\prime}(x) d x+c$
  • D
    $\left(\psi_0 \phi\right)(x) h(x)-\int\left(\psi_0 \phi\right)(x) h^{\prime}(x) d x+c$

Solution

$\int \phi(x) d x=\psi(x)$, then $\int(\phi \circ h)(x) \cdot h(x) h^{\prime}(x) d x=$ ? Now, $\int \phi\{h(x)\} \cdot h(x) \cdot h^{\prime}(x) d x$ Let $h(x)=t \Rightarrow h^{\prime}(x) d x=d t$ Now, $\int_{\mathrm{II}}^{\phi(t) \cdot{ }_{\mathrm{I}}} d t$ $=t \cdot \int \phi(t) d t-\int\left\{\frac{d}{d t}(t) \cdot \int \phi(t) d t\right] d t=t \cdot \psi(t)-\int \psi(t) d t$ $=h(x) \psi_0 h(x)-\int \psi_0 h(x) \cdot h^{\prime}(x) \cdot d x$

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

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