Trapezium Rule

**Hellbent** · November 8th 2010, 05:39 PM

Hi,

I need help checking these, please.

i.) Use the trapezium rule with intervals of width $0.5$ to find an approximation for $\int_1^{2.5} \frac{1}{1\,+\,\ln x}\,dx$

$h = 0.5$

$\begin{vmatrix} x_r & 1 & 1.5 & 2.0 & 2.5 \\y_r & 1 & 0.7115 & 0.5906 & 0.5218 \end{vmatrix}$

$\int_1^{2.5} \frac{1}{1\,+\,\ln x}dx = \frac{1}{2}\cdot0.5\{1+2(0.7115+0.5906)\,+\,0.5218 \}$

$\approx 1.0315$

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

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

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

$h = 0.4$

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

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

$\approx 6.961$

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

$\approx 5.96791$ How can this be expressed in terms of e?

**Unknown008** · November 8th 2010, 11:33 PM

1. Okay.

2. Right.

For this, I'll use a substitution of x^2 = u

du = 2x dx

dx = 1/2x du

u = 1.6^2 = 2.56
u = 0^2 = 0

$\begin{array}{rcl} \displaystyle \int^{1.6}_0 xe^{x^2}\ dx &=&\displaystyle \int^{2.56}_0 \frac{xe^u}{2x}\ du \\ && \\ &=&\displaystyle \dfrac12 \int^{2.56}_0 e^u\ du \\ && \\ &=& \dfrac12 \left[e^u\right]^{2.56}_0 \\ && \\ &=& \dfrac12 \left[e^{2.56} - 1\right] \end{array}$

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