# How do you calculate the value of the integral inte^(4t²-t) dt from [3,x]?

Aug 1, 2015

$\int {e}^{4 {t}^{2} - t} \mathrm{dt} = \frac{{e}^{4 {x}^{2} - x}}{8 x - 1} - {e}^{33} / 23$

#### Explanation:

Be $f \left(x\right) = {e}^{4 {t}^{2} - t}$ your function.

In order to integrate this function, you will need its primitive $F \left(x\right)$

$F \left(x\right) = \frac{{e}^{4 {t}^{2} - t}}{8 t - 1} + k$ with $k$ a constant.

The integration of ${e}^{4 {t}^{2} - t}$ on [3;x] is calculated as follows:

$\int {e}^{4 {t}^{2} - t} \mathrm{dt} = F \left(x\right) - F \left(3\right)$

$= \frac{{e}^{4 {x}^{2} - x}}{8 x - 1} + k - \left(\frac{{e}^{4 \cdot {3}^{2} - 3}}{8 \cdot 3 - 1} + k\right)$

$= \frac{{e}^{4 {x}^{2} - x}}{8 x - 1} - {e}^{33} / 23$

Aug 1, 2015

That integral cannot be expressed using elementary functions. If requires the use of $\int {e}^{{x}^{2}} \mathrm{dx}$. However the derivative of the integral is ${e}^{4 {x}^{2} - x}$

#### Explanation:

The fundamental theorem pf calculus part 1 tells us that the derivative with respect to $x$ of:

$g \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$ is $f \left(x\right)$

So the derivative (with respect to $x$) of

$g \left(x\right) = {\int}_{3}^{x} {e}^{4 {t}^{2} - t} \mathrm{dt} \text{ }$ is $\text{ } g ' \left(x\right) = {e}^{4 {x}^{2} - x}$.