Originally Posted by

**Grandad** Hi -

Here's some info that should help:

$\displaystyle \int\text{ln }u.du=u\text{ ln }u-u+c$

(If you want to check this, just differentiate the RHS using the product rule.)

You were given that

$\displaystyle \int u\text{ ln } u.du=\frac{u^2}{4}(2\text{ ln }u-1)+c$

and you need to find:

$\displaystyle I=\int_0^{25}t \text{ ln }(1+0.53t).dt$

before dividing the answer by 25, to get the average depth of snow.

To find the value of the integral *I*, use the substitution:

$\displaystyle 1+0.53t=u$

So: $\displaystyle t=\frac{u-1}{0.53}$

and: $\displaystyle dt = \frac{du}{0.53}$

When you do the substitution, replace the limits of the integral by the corresponding values of u. Thus:

$\displaystyle I=\int_1^{14.25}\frac{u-1}{0.53}\text{ ln }u.\frac{du}{0.53}$

**You'll see now that you've got to integrate $\displaystyle (u-1)\text{ ln }u$.**

Do this as two separate terms: $\displaystyle u\text{ ln }u$ and $\displaystyle -\text{ ln }u$, using the integral formulae above.

If I've done the arithmetic correctly, the average depth comes out to be about 27.7 cm (but it would be good if you could check that!).

I hope that helps.

Grandad