y = 5^x + ext{log}_e(x + 1), ext{ } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } 2 Complete the table below, by giving the value of $y$ when $x = 1$ | $x$ | $0$ | $0.5$ | $1.5$ | $2$ | |-----|-----|-------|-------|-----| | $y$ | $1$ | $2.821$ | $12.502$ | $26.585$ | (b) Use the trapezium rule, with all the values of $y$ from the completed table, to find an approximate value for $$\int_0^2 (5^x + ext{log}_e(x + 1))dx$$ giving your answer to 2 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 2

Using the trapezium rule:

the formula is: $ext{Area} \approx \frac{b-a}{2}(f(a) + f(b)) + \sum_{i=1}^{n-1} f(x_i)$

Here, $a = 0$, $b = 2$, and $y$ values are as calculated above. There are 4 intervals, each of width $h = 0.5$:

$ext{Area} \approx \frac{0.5}{2} ((1) + (26.585)) +\frac{1}{2}(2.821 + 12.502)\n= 0.25(1 + 26.585) + 0.5(2.821 + 12.502)$

Calculating this gives:

• First part: $0.25 \times 27.585 \approx 6.89625$,
• Second part: $0.5 \times 15.323 \approx 7.6615$.

Thus: $\text{Total Area} \approx 6.89625 + 7.6615 \approx 14.55775 \approx 14.56$

Hence, the approximate value is $14.56$.

In part (b), we found the approximate value for: $\int_0^2 (5^x + \text{log}_e(x + 1))dx \approx 14.56$

Now, we want to find: $\int_0^2 (5 + 5^x + \text{log}_e(x + 1))dx = \int_0^2 5dx + \int_0^2 (5^x + \text{log}_e(x + 1))dx$

The first integral is: $\int_0^2 5dx = 5 \times 2 = 10.$

Therefore: $\int_0^2 (5 + 5^x + \text{log}_e(x + 1))dx \approx 10 + 14.56 = 24.56.$

So, we approximate: $24.56 \approx 24.56.$