How do you find the area using the trapezoidal approximation method, given e^(x^2), on the interval [0,1] with n=10?

Jan 5, 2017

${\int}_{0}^{1} {e}^{{x}^{2}} \setminus \mathrm{dx} \approx 1.4672$ (4dp)

Explanation:

The values of $f \left(x\right) = {e}^{{x}^{2}}$ are tabulated as follows (using Excel) working to 6dp. The Trapezium Rule:

${\int}_{a}^{b} y \mathrm{dx} \approx \frac{h}{2} \left\{\left({y}_{0} + {y}_{n}\right) + 2 \left({y}_{1} + {y}_{2} + \ldots + {y}_{n - 1}\right)\right\}$

uses a series of two consecutive ordinates and a best fit straight line to form trapeziums to approximate the area under a curve, It will have 100% accuracy if $y = f \left(x\right)$ is a straight line and typically provides a good estimate provided $n$ is chosen appropriately.

So,

 int_0^1 e^(x^2) \ dx ~~ 0.1/2 { (1 + 2.718281) +
 " " 2(1.01005 + 1.04081 + 1.094174 + 1.17351 + 1.284025 +
 " " 1.433329 + 1.632316 + 1.89648 + 2.247907) }
$\text{ } = 0.05 \left\{+ 3.718281 + 2 \left(12.812606\right)\right\}$
$\text{ } = 0.05 \left\{+ 3.718281 + 25.625212\right\}$
$\text{ } = 0.05 \left\{+ 29.343493\right\}$
$\text{ } = 1.467174$