Using u-subsitution to integrate $ \int_{0}^{1} e^{3 x^{3}-2 x}\left(9 x^{2}-2\right) d x $

Question

Question: Use the u-subsitution to integrate $ \int_{0}^{1} e^{3 x^{3}-2 x}\left(9 x^{2}-2\right) d x $

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So far my working is as follows

\begin{equation} u=3 x^{3}-2 x \Longrightarrow \frac{d u}{d x}=9 x^{2}-2 \end{equation} Step two:

\begin{equation} \frac{d x}{d u}=\frac{1}{9 x^{2}-2} \end{equation}

Substituting u in

\begin{equation} \int_{0}^{1} e^{u}\left(9 x^{2}-2\right) d x \end{equation}

For definite integration using substitution what changes occur as compared to indefinite u substitution?

EDIT:

After looking at this question I believe I am close to the answer

\begin{equation} \int_{0}^{1} e^{4}\left(9 x^{2}-2\right) \frac{1}{9 x^{2}-2} d x \end{equation}

Canceling them out we get;

\begin{equation} \int_{0}^{1} e^{u} d u \end{equation}

when x=1, our new limit is 7 when x=0 our new limit is -2

Answer 1

Your expression after u-substitution is just $$\int_0^1 e^u du=e^{3x^2-2x}\Bigg |^1_0$$ You can do the rest.

Answer 2

\begin{equation} \frac{d x}{d u}=\frac{1}{9 x^{2}-2}, \end{equation}

We are allowed to multiply both sides by $du$ to get:

\begin{equation} d x=\frac{1}{9 x^{2}-2} du. \end{equation}

Substitute this into \begin{equation} \int_{x=0}^{x=1} e^{u}\left(9 x^{2}-2\right) d x. \end{equation}

But then you must get the limits $x=1$ and $x=0$ also in terms of $u.$

So you have to replace $x=1$ with $u=?$ [What is $u$ when $x=1$?]

And similar for $x=0.$

Only once everything is in terms of $u$ are you allowed to evaluate the integral.

Answer 3

So from my workings I computed

\begin{equation} \left(e^{u}\right)_{0}^{1}=e^{1}-e^{0}=e-1 \end{equation}

This is from

\begin{equation} \int_{0}^{1} e^{u} d u \end{equation}

Therefore giving our final answer

\begin{equation} \left(e^{u}\right)_{0}^{1}=e^{1}-e^{0}=e-1 \end{equation}

Answer 4

Upon inspection you can write $$ \int_0^1e^{3x^3-2x}(9x^2-2)\,\mathrm dx=\int_0^1e^{3x^3-2x}\,\mathrm d(3x^3-2x). $$ Since we are now integrating with respect to the quantity in the exponent, the result follows immediately $$ \int_0^1e^{3x^3-2x}\,\mathrm d(3x^3-2x)=e^{3x^3-2x}|_{x=0}^1=e-1. $$ Admittedly, this is just $u$-substitution in disguise.