Let the equation of the curve passing through the point \((0,1)\) be given by \(y=\int x^3 e^{x^4} d x\). If…

Let the equation of the curve passing through the point \((0,1)\) be given by \(y=\int x^3 e^{x^4} d x\). If the equation of the curve is written in the form \(x=f(y)\), then \(f(y)=\)
  1. \(\log |4 y-3|\)
  2. \((\log |4 y-3|)^{1 / 4}\)
  3. \(\left(\log \left|\frac{3-4 y}{4}\right|\right)^{1 / 4}\)
  4. \(\log \left|\frac{4 y-3}{4}\right|\)

Solution

It is given that, \(y=\int x^3 e^{x^4} d x\) Let \(x^4=t \Rightarrow 4 x^3 d x=d t\) \(\therefore\) So, \(y=\frac{1}{4} \int e^t d t=\frac{e^t}{4}+C=\frac{e^{x^4}}{4}+C\) \(\because\) The curve \(y=\frac{e^{x^4}}{4}+C\) passes through point \((0,1)\), so \(C=\frac{3}{4} \Rightarrow y=\frac{e^{x^4}}{4}+\frac{3}{4} \Rightarrow e^{x^4}=4 y-3\) \(\Rightarrow \quad x^4=\log _e|4 y-3| \Rightarrow x=\left(\log _e|4 y-3|\right)^{1 / 4}\) Hence, option (c) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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