Pearson Edexcel IAL Mathematics P2.8 Integration Question Bank

Practise evaluating integrals, interpreting them as areas, and using trapezium-rule estimates from tables or transformed integral information.

Exam points

evaluate definite integrals by substituting limits into an antiderivative
find bounded areas using curve integrals, straight-line areas, or differences of curves
apply the trapezium rule with table values and state an estimate to the required accuracy

Covered syllabus points

P2.8.1 - Evaluation of definite integrals P2.8.2 - Interpretation of the definite P2.8.3 - Trapezium rule for area approximation

[Maximum number: 6]

(a)

Table for Question (a) — Edexcel A-Level Mathematics AS

The table shows corresponding values of x and y for

y=2^x-2x

with the values of y given to 4 decimal places as appropriate.
Using the trapezium rule with all the values of y in the given table,
obtain an estimate for $\int_{2}^{3.5}(2^x-2x)\,dx$ , giving your answer to 2 decimal places.

[ 3 ]

B1: For h=0.3 This is implied by sight of $\frac{0.3}{2}$ in front of the bracket.
M1: Applies the trapezium rule with correct bracket condoning slips in copying (or even truncating) values. May use separate trapezia. Allow a missing final bracket, but otherwise bracketing must be correct or recovered by correct answer.
A1: awrt 2.34 Note that the calculator answer for this integral is 2.30

(b)

Using your answer to part (b) and making your method clear, estimate

[ 3 ]

(i)

$\int_{2}^{3.5}\left(2^{x}+2 x\right) \mathrm{d} x$

[ 2 ]

$\int_2^{3.5}(2^x+2x)\,dx=\int_2^{3.5}(2^x-2x+4x)\,dx$ $=2.34+[2x^2]_2^{3.5}$ =2.34+16.5=18.84

M1
A1ft

M1: Realises that the integral can be written as $\int_2^{3.5}(2^x-2x+4x)\,dx$ and integrates the 4x term.
A1ft: 2.34+16.5=18.84, following through on the answer to part (b); allow awrt 3 s.f.

(ii)

$\int_{2}^{3.5}\left(2^{x+1}-4 x\right) \mathrm{d} x$

[ 1 ]

B1 ft: For $2\times$ their answer to part (b), so $2\times2.34=4.68$ (awrt 3 s.f.; must be evaluated).
Note: Answer via repeated trapezium rule is permitted to score the B1 as long as the answer is twice their part (b).

[Maximum number: 3]

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

Figure 3

Figure 3 shows
- the curve $C_{1}$ with equation $y=x^{3}-5 x^{2}+3 x+14$
- the circle $C_{2}$ with centre T

The point T is the minimum turning point of $C_{1}$
Using Figure 3 and calculus,

Find the exact area of R.

$\int_0^2 (x^3-5x^2+3x+14)\,dx =\left[\frac14x^4-\frac53x^3+\frac32x^2+14x\right]_0^2$
The triangular area to subtract is
$\frac12\left(\frac{22}{3}+8\right)2=\frac{46}{3}.$
Hence
$\text{Area }R=\frac{74}{3}-\frac{46}{3}=\frac{28}{3}.$
M1 dM1 A1

[Maximum number: 3]

find an approximate value for

\int_{-4}^{8.5}\left(3^{-\frac{1}{2} x}+4\right) d x

giving your answer to two significant figures.

Strip width =2.5.
$\int_{-4}^{8.5}\left(3^{-\frac12x}+4\right)\,dx \approx \frac54\{13+4.009+2(6.280+4.577+4.146+4.037)\}$ $=68.86125\approx69.$
B1 M1 A1

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Covered syllabus points

P2.8 - Integration question 1

Question (a)

Question (b)

Question (i)

Question (ii)

P2.8 - Integration question 2

P2.8 - Integration question 3