If $ g'(x^2) = x^3$ for all $x > 0$ and $g(1) = 1$ then we can find $g(x)$ by integrating $\int g'(y) = \int xy $ assuming $x^2 = y$. Proceeding with the integration with respect to variable $y$, we would get $g(y) = \frac{x{y^2}}{2}$. How to proceed further?
Let $g :\mathbb{R} \rightarrow \mathbb{R} $ be differentiable with $ g'(x^2) = x^3$ for all $x > 0$ and $g(1) = 1$. Then $g(4)$ equals?
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integration
derivatives
2 Answers
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With $g'(x^2).2x=2x^4$ then $$\int_1^t g'(x^2)2x.dx=\int_1^t 2x^4.dx$$ and $g(t^2)=\dfrac25t^5+\dfrac35$ so $g(4)=\dfrac{67}{5}$.
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Your integration of $xy$ does not make sense because $x$ and $y$ are not independent, i.e. $x$ is not constant with respect to $y$. Actually you are integrating $xy = x^3 = y^{\frac{3}{2}}$.
Other than that the approach by substitution is good. Since $g'(x^2) = x^3$ for all $x>0$, it is justified to write $g'(y) = y^{\frac{3}{2}}$ for all $y > 0$ (for negative values of $y$ this substitution would no longer make sense, given the square root).
Now you can integrate both sides with respect to $y$, as you wanted, but don't forget the constant that comes from indefinite integration! You should get $\int g'(y) \,dy = \int y^{\frac{3}{2}} \,dy$. You should then get $g(y) = \frac{2}{5}y^{\frac{5}{2}} + C$
Now you can use your initial condition and substitution to find $C$ and then find $g(4)$.