TMUA: Integration: the Fundamental Theorem and the trapezium rule

Integrating powers, the Fundamental Theorem, combining integrals and the trapezium rule.

Section 1 Part 1 · MM7 Integration · spec MM7.3-MM7.6
Papers 1 and 2
9 practice questions
No calculator

Practise integration: the fundamental theorem and the trapezium rule →Timed TMUA paper

What the specification covers

F(b) - F(a); d/dx of an integral with variable upper limit
Combining ranges and integrands
Trapezium rule over- or underestimate from convexity
dy/dx = f(x)

Key ideas

\(\displaystyle\int x^n\,dx=\frac{x^{n+1}}{n+1}+c\) for \(n\ne-1\); simplify before integrating.
\(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)\).
\(\int_a^b f+\int_b^c f=\int_a^c f\) and \(\int_a^b(f+g)=\int_a^b f+\int_a^b g\).
The trapezium rule overestimates where the curve bends upwards (convex) and underestimates where it bends downwards.

Common mistakes

Integrate each term of \(\dfrac{(x+1)^2}{\sqrt x}\) after expanding and dividing, not before.
The trapezium rule width is \(h=\tfrac{b-a}{n}\) for \(n\) strips (\(n+1\) ordinates).

Exam tip

For an unknown limit, the integral gives a polynomial equation in that limit; use the given condition (such as \(a>0\)) to choose the root.

Worked example

Evaluate \(\displaystyle\int_1^4\frac{(x+1)^2}{\sqrt{x}}\,dx\).

\(24\)
\(\frac{412}{15}\)
\(\frac{356}{15}\)
\(\frac{178}{15}\)
\(\frac{326}{15}\)

Answer: C: \(\frac{356}{15}\)

\(\dfrac{(x+1)^2}{\sqrt x}=x^{3/2}+2x^{1/2}+x^{-1/2}\), with antiderivative \(\tfrac25x^{5/2}+\tfrac43x^{3/2}+2x^{1/2}\). From 1 to 4: \(\tfrac25(32-1)+\tfrac43(8-1)+2(2-1)=\tfrac{62}{5}+\tfrac{28}{3}+2=\tfrac{356}{15}\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-3)\,dx=4\), what is \(a\)?

\(3\)
\(\frac{3}{2}\)
\(4\)
\(1\)
\(2\)

Show the answer and solution

Answer: C: \(4\)

\(\int_0^a(2x-3)\,dx=a^2-3a=4\), so \((a-4)(a+1)=0\). As \(a>0\), \(a=4\).

Question 2

The trapezium rule with 2 strips of equal width is used to estimate \(\displaystyle\int_0^2 2^x\,dx\). Which of the following is correct?

The estimate is \(4\), an underestimate
The estimate is \(9\), an overestimate
The estimate is \(4\), an overestimate
The estimate is \(4.5\), an overestimate
The estimate is \(4.5\), an underestimate

Show the answer and solution

Answer: D: The estimate is \(4.5\), an overestimate

Strip width 1, ordinates \(1,2,4\): estimate \(\tfrac12(1+2\cdot2+4)=4.5\). The graph of \(2^x\) curves upwards (convex), so each chord lies above the curve and the estimate is too large.

Question 3

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-3)\,dx=18\), what is \(a\)?

\(3 \sqrt{2}\)
\(-3\)
\(9\)
\(3\)
\(6\)

Show the answer and solution

Answer: E: \(6\)

\(\int_0^a(2x-3)\,dx=a^2-3a=18\), so \((a-6)(a+3)=0\). As \(a>0\), \(a=6\).

Question 4

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-1)\,dx=20\), what is \(a\)?

\(5\)
\(-4\)
\(2 \sqrt{5}\)
\(1\)
\(6\)

Show the answer and solution

Answer: A: \(5\)

\(\int_0^a(2x-1)\,dx=a^2-1a=20\), so \((a-5)(a+4)=0\). As \(a>0\), \(a=5\).

Question 5

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-1)\,dx=30\), what is \(a\)?

\(-5\)
\(\sqrt{30}\)
\(7\)
\(6\)
\(1\)

Show the answer and solution

Answer: D: \(6\)

\(\int_0^a(2x-1)\,dx=a^2-1a=30\), so \((a-6)(a+5)=0\). As \(a>0\), \(a=6\).

Question 6

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-1)\,dx=12\), what is \(a\)?

\(2 \sqrt{3}\)
\(4\)
\(-3\)
\(1\)
\(5\)

Show the answer and solution

Answer: B: \(4\)

\(\int_0^a(2x-1)\,dx=a^2-1a=12\), so \((a-4)(a+3)=0\). As \(a>0\), \(a=4\).

Question 7

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-2)\,dx=24\), what is \(a\)?

\(2\)
\(2 \sqrt{6}\)
\(6\)
\(-4\)
\(8\)

Show the answer and solution

Answer: C: \(6\)

\(\int_0^a(2x-2)\,dx=a^2-2a=24\), so \((a-6)(a+4)=0\). As \(a>0\), \(a=6\).

Question 8

Given that \(a>0\) and \(\displaystyle\int_0^a(2x-1)\,dx=6\), what is \(a\)?

\(3\)
\(4\)
\(-2\)
\(\sqrt{6}\)
\(1\)

Show the answer and solution

Answer: A: \(3\)

\(\int_0^a(2x-1)\,dx=a^2-1a=6\), so \((a-3)(a+2)=0\). As \(a>0\), \(a=3\).

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