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if $\displaystyle g(x) = \int^{x}_{1}e^{t^2} dt,$ then $\displaystyle \int^{x^3}_{3}e^{t^2}dt$ in terms of $g(x)$

Attempt: from $\displaystyle g(x) = \int^{x}_{1}e^{t^2} dt$, we have $\displaystyle g'(x) = e^{x^2}$ and

let $\displaystyle h(x) = \int^{x^3}_{3}e^{t^2} dt,$ we have $\displaystyle h'(x) = e^{x^6}\cdot 3x^2$

could some help me to solve it, thanks

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    What is $g(x^3)$? You can use the fact that $\int_{a}^{c} = \int_{a}^{b} + \int_{b}^{c}$ – 2017-02-27
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    I believe you made a mistake with the derivative of $h(x)$ as it should be $h'(x)=3x^2e^{x^6}$ – 2017-02-27

2 Answers 2

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$ h(x) = \int^{x^3}_{3}e^{t^2} dt=\int^{x^3}_{1}e^{t^2}dt -\int^{3}_{1}e^{t^2} dt=g(x^3)-g(3)$

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$$\int_3^{x^3}e^{t^2}dt=\int_3^1e^{t^2}dt+\int_1^{x^3}e^{t^2}dt=-g(3)+g(x^3).$$

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    Thats not o.k. Better. $\int_3^{x^3}e^{t^2}dt=-\int_1^3e^{t^2}dt+\int_3^{x^3}e^{t^2}dt=-g(3)+g(x^3).$ – 2017-02-27
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    In the first equality you have added $\int_{1}^{3}{e^{t^{2}}}$ out of nowhere. It should be $g(x^3) - g(3)$ – 2017-02-27
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    Fixed the typo. – 2017-02-27