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!