Hello headbazza

The simplest method of doing a numerical integration is to use the Trapezium Rule.

Imagine the area under the curve $\displaystyle y = f(x)$ divided vertically into $\displaystyle n$ strips of equal width $\displaystyle d$, with the curve replaced by straight line segments. Then the area is approximately the same as the sum of the areas of the trapezia; which is:$\displaystyle \Big(\frac{f(x_0)+f(x_n)}{2} + f(x_1) + f(x_2) + ... + f(x_{n-1})\Big)d$, where $\displaystyle x_0, x_1, ...,x_n$ are the $\displaystyle n+1$ equally spaced values of $\displaystyle x$.

So, to use this formula in your question, you'll work out $\displaystyle y(0), y(10), y(20), ... ,y(60)$, and then:Area = $\displaystyle 10\Big(\frac{y(0)+y(60)}{2} + y(10) + y(20) + ... + y(50)\Big)$

Take care with the calculation, and it should be very straightforward.

For (b):Area $\displaystyle = \int_0^{60}e^{0.06t}dt$$\displaystyle = \frac{1}{0.06}\Big[e^{0.06t}\Big]_0^{60}$

$\displaystyle = \frac{1}{0.06}(e^{3.6}-e^0)$

$\displaystyle \approx 593$

For (c), the answer to (b) will be more accurate because in (a) the curve has been replaced by straight line segments joining the ordinates at $\displaystyle x = 0, 10, ..., 60$.

Grandad