Let the equation of the curve passing through | vedclass

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(B) Given the equation of the curve is $y = \int x^3 e^{x^4} dx$. Let $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substitut

Vedclass · Accepted Answer

$(\log |4 y-3|)^{1 / 4}$

Vedclass · Answer

(B) Given the equation of the curve is $y = \int x^3 e^{x^4} dx$. Let $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting this into the integral,we get $y = \frac{1}{4} \int e^t dt = \frac{1}{4} e^t + C = \frac{1}{4} e^{x^4} + C$. Since the curve passes through $(0,1)$,we substitute $x=0$ and $y=1$: $1 = \frac{1}{4} e^0 + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$. Thus,the equation is $y = \frac{1}{4} e^{x^4} + \frac{3}{4}$. Rearranging for $x$: $y - \frac{3}{4} = \frac{1}{4} e^{x^4} \Rightarrow 4y - 3 = e^{x^4}$. Taking the natural logarithm on both sides: $\log_e |4y - 3| = x^4$. Therefore,$x = (\log_e |4y - 3|)^{1/4}$. Thus,$f(y) = (\log_e |4y - 3|)^{1/4}$.

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 $f(y)=$

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