The table below shows corresponding values of x and y for y = log_2 x The values of y are given to 2 decimal places as appropriate - Edexcel - A-Level Maths Pure - Question 7 - 2022 - Paper 2

Using the result from part (a), we can express the integral as:

$$\int_3^9 log_2 (2x)^{10} \, dx = 10 \int_3^9 log_2 2x \, dx.$$

Using the earlier estimate for ( \int_3^9 log_2 2x , dx ), we find:

$$10 \times 10.35 = 103.5.$$

Thus, the estimate for ( \int_3^9 log_2 (2x)^{10} , dx ) is approximately 103.5.

To estimate ( \int_3^9 log_2 18x , dx ), we can rewrite it as:

$$\int_3^9 log_2 18 + log_2 x \, dx.$$

This leads to:

$$\int_3^9 log_2 18 \, dx + \int_3^9 log_2 x \, dx.$$

The first integral is:

$$log_2 18 \times (9 - 3) = log_2 18 \times 6.$$

Calculating this gives:

$$6 \times 4.1699 = 25.0194 \\ \text{(using } log_2 18 \approx 4.1699 \text{)}.$$

Next, for ( \int_3^9 log_2 x , dx ), we again use the trapezium rule as in part (a) with values:

• ( y = 1.63, 2.26, 2.46, 2.63 )

The estimate will be the same as before, yielding 10.35. Hence, we have:

$$25.02 + 10.35 = 35.37.$$

Thus, the estimate for ( \int_3^9 log_2 18x , dx ) is approximately 35.37.