Trapezoidal Rule (Verification)

[Erik Lehnsherr]

Just need verification.

$\displaystyle \text{i.\,\,\,\,Use the trapezium rule with intervals of width}$ $\displaystyle \,0.4$ $\displaystyle \text{to find an approximate value for}$ $\displaystyle \,\,\int_0^{1.6} xe^{x^2}\,dx$

$\displaystyle \int_0^{1.6} xe^{x^2} dx$

$\displaystyle h = 0.4$

$\displaystyle \begin{vmatrix} x_r & 0 & 0.4 & 0.8 & 1.2 & 1.6 \\y_r & 0 & 0.4694 & 1.517 & 5.065 & 20.70 \end{vmatrix}$

$\displaystyle \int_0^{1.6} xe^{x^2} dx = \frac{1}{2}\cdot0.4\{2(0.4694+1.517+5.065)+20.70\}$

$\displaystyle \approx 6.961$

$\displaystyle \text{ii\,\,\,\,Calculate the exact value of}$ $\displaystyle \int_0^{1.6} xe^{x^2}\,dx$ $\displaystyle \text{leaving your answer in terms of e.}$

$\displaystyle \int_0^{1.6} xe^{x^2} dx$

$\displaystyle \approx 5.96791$

 

[BigGlenntheHeavy]

$\displaystyle \int_{0}^{1.6} x e^{x^{2}}dx, \ let \ u \ = \ x^2 \ \implies \ du \ = \ 2xdx \ \implies \ \frac{du}{2} \ = \ xdx.$

$\displaystyle Hence, \ we \ have: \ \frac{1}{2}\int_{0}^{2.56} e^udu \ = \ \frac{e^u}{2}\bigg]_{0}^{2.56} \ = \ \frac{e^{2.56}-1}{2}.$

 

[Erik Lehnsherr]

Gracias!