Complete the table below, by giving the value of $y$ when $x = 1$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 2
Question 5
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... show full transcript
Worked Solution & Example Answer:Complete the table below, by giving the value of $y$ when $x = 1$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 2
Step 1
Complete the table below, by giving the value of $y$ when $x = 1$
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the value of y when x=1, we substitute x=1 into the equation:
y=51+log(1+1)=5+log(2)≈5+0.301=5.301.
Thus, the completed table looks like:
x
0
0.5
1
1.5
2
y
1
2.821
5.301
12.502
26.585
Step 2
Use the trapezium rule, with all the values of $y$ from the completed table, to find an approximate value for $\int_0^{2} (5 + \log(x + 1)) \;dx$
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using the trapezium rule:
We have width h=0.5.
The area can be approximated as:
T=2h(y0+2y1+2y2+2y3+y4) T=20.5(1+2⋅2.821+2⋅5.301+2⋅12.502+26.585) T=0.25⋅(1+5.642+10.602+25.004+26.585) T=0.25⋅68.833=17.20825.
Rounding to 2 decimal places gives an approximate value of 17.21.
Step 3
Use your answer to part (b) to find an approximate value for $\int_0^{2} (5 + 5 \cdot \log(x + 1)) \;dx$
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
From part (b), we approximate:
∫02(5+log(x+1))dx≈17.21.
Thus,
∫02(5+5⋅log(x+1))dx≈5⋅∫02(5+log(x+1))dx =5imes17.21=86.05.
Rounding to two decimal places gives an approximate value of 86.05.
