Integration | The Maths Tailor — UK Methods

Integration

9 methods · 34 questions

A definite integral gives a signed area, so a question asking for the total area enclosed is not a single integral. Find every point where the curve meets the $x$ -axis (or where $∣ x ∣$ switches branch), integrate over each sub-interval separately, then add the absolute values so the parts below the axis count positively.

Trigger: the words 'total area' or 'area enclosed', a curve that crosses the axis, or an integrand containing $∣ x ∣$ .

Instances:
(i) factorise to locate the roots, then sum $∣ \int ∣$ over the sign-constant pieces;
(ii) split the range at $x = 0$ when $∣ x ∣$ appears so each branch becomes an ordinary polynomial;
(iii) exploit symmetry about the $y$ -axis to integrate one side and double it.

Linked questions (5)

TMUA 2016 Paper 1 — Q5

What is the total area enclosed between the curve

y = x^{2} - 1

, the

x

-axis and the lines

x = - 2

and

x = 2

?

(A)

\frac{4}{3}

(B)

\frac{8}{3}

(C)

4

(D)

\frac{16}{3}

(E)

12

(F)

16

Voir la correction

The parabola

y = x^{2} - 1 = (x + 1) (x - 1)

crosses the

x

-axis at

x = \pm 1

. The curve is below the

x

-axis on

(- 1, 1)

and above it on

(- 2, - 1)

and

(1, 2)

. Compute the three signed integrals:

\int_{- 2}^{- 1} (x^{2} - 1) d x = [\frac{x ^{3}}{3} - x]_{- 2}^{- 1} = (- \frac{1}{3} + 1) - (- \frac{8}{3} + 2) = \frac{2}{3} - (- \frac{2}{3}) = \frac{4}{3}

.

\int_{- 1}^{1} (x^{2} - 1) d x = [\frac{x ^{3}}{3} - x]_{- 1}^{1} = (\frac{1}{3} - 1) - (- \frac{1}{3} + 1) = - \frac{2}{3} - \frac{2}{3} = - \frac{4}{3}

. By symmetry,

\int_{1}^{2} (x^{2} - 1) d x = \frac{4}{3}

. Total area

= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4

. Hence the answer is C.

TMUA 2018 Paper 1 — Q12

A curve has equation

y = f (x)

, where

f (x) = x (x - p) (x - q) (r - x)

with

0 < p < q < r

. You are given that:

\int_{0}^{r} f (x) d x = 0, \int_{0}^{q} f (x) d x = - 2, \int_{p}^{r} f (x) d x = - 3.

What is the total area enclosed by the curve and the

x

-axis for

0 \leq x \leq r

?

(A)

0

(B)

1

(C)

4

(D)

5

(E)

6

(F)

10

Voir la correction

The curve

f (x) = x (x - p) (x - q) (r - x)

has roots at

0, p, q, r

and leading term

- x^{4}

, so it is positive on

(0, p)

, negative on

(p, q)

, positive on

(q, r)

... wait, actually for a degree-4 polynomial with negative leading coefficient, it tends to

- \infty

at both ends. Checking signs: for small positive

x

, all factors except

(r - x)

are small and positive,

(r - x) > 0

, so

f (x) > 0

on

(0, p)

. For

x \in (p, q)

:

(x - p) > 0

,

(x - q) < 0

,

(r - x) > 0

, so

f < 0

. For

x \in (q, r)

: all differences positive except

(r - x) > 0

and the product

x (x - p) (x - q) > 0

, but

(r - x) > 0

, actually here

(x - p) > 0

,

(x - q) > 0

, so the product of all factors is

x (x - p) (x - q) (r - x) > 0

on

(q, r)

... let me re-examine. Actually on

(q, r)

:

x > 0

,

(x - p) > 0

,

(x - q) > 0

,

(r - x) > 0

, so

f > 0

. So

f

is positive on

(0, p)

, negative on

(p, q)

, positive on

(q, r)

. Label areas:

A = \int_{0}^{p} f d x > 0

,

B = - \int_{p}^{q} f d x > 0

,

C = \int_{q}^{r} f d x > 0

. Then: (1)

\int_{0}^{r} = A - B + C = 0

; (2)

\int_{0}^{q} = A - B = - 2

; (3)

\int_{p}^{r} = - B + C = - 3

. From (1) and (3):

A = (A - B + C) - (- B + C) = 0 - (- 3) = 3

. From (2):

B = A + 2 = 5

. From (1):

C = B - A = 2

. Total area

= A + B + C = 3 + 5 + 2 = 10

. Hence the answer is F.

TMUA 2019 Paper 1 — Q10

Evaluate

\int_{- 1}^{3} ∣ x ∣ (1 - x) d x

(A)

\frac{17}{3}

(B)

- \frac{17}{3}

(C)

\frac{16}{3}

(D)

- \frac{16}{3}

(E)

\frac{11}{3}

(F)

- \frac{11}{3}

Voir la correction

Split at

\frac{11}{3}

where

∣ x ∣

changes sign. For

- 1 \leq x \leq 0

,

∣ x ∣ = - x

; for

0 \leq x \leq 3

,

∣ x ∣ = x

.

\int_{- 1}^{0} (- x) (1 - x) d x + \int_{0}^{3} x (1 - x) d x

= \int_{- 1}^{0} (- x + x^{2}) d x + \int_{0}^{3} (x - x^{2}) d x

= [- \frac{x ^{2}}{2} + \frac{x ^{3}}{3}]_{- 1}^{0} + [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3}

= (0 - (- \frac{1}{2} - \frac{1}{3})) + (\frac{9}{2} - 9) = \frac{5}{6} + (- \frac{9}{2}) = \frac{5}{6} - \frac{27}{6} = - \frac{22}{6} = - \frac{11}{3} .

Hence the answer is F.

TMUA 2021 Paper 1 — Q7

The function f is such that

f (0) = 0

, and

x f (x) > 0

for all non-zero values of

x

. It is given that

\int_{- 2}^{2} f (x) d x = 4 and \int_{- 2}^{2} ∣ f (x) ∣ d x = 8.

Evaluate

\int_{- 2}^{0} f (∣ x ∣) d x

(A)

- 8

(B)

- 6

(C)

- 4

(D)

- 2

(E)

2

(F)

4

(G)

6

(H)

8

Voir la correction

From

x f (x) > 0

for all non-zero

x

:

f (x) > 0

for

x > 0

and

f (x) < 0

for

x < 0

. Let

A = \int_{- 2}^{0} f (x) d x

and

B = \int_{0}^{2} f (x) d x

. Then

A + B = 4

. Since

∣ f (x) ∣ = - f (x)

for

x < 0

and

∣ f (x) ∣ = f (x)

for

x > 0

:

- A + B = 8

. Adding:

2 B = 12

, so

B = 6

and

A = - 2

. Now

\int_{- 2}^{0} f (∣ x ∣) d x

: the graph of

y = f (∣ x ∣)

for

x \in [- 2, 0]

is the reflection of

y = f (x)

for

x \in [0, 2]

in the

y

-axis, so the integral equals

\int_{0}^{2} f (x) d x = 6

. Hence the answer is G.

TMUA 2022 Paper 1 — Q7

Find the finite area enclosed between the line

y = 0

and the curve

y = x^{2} - 4∣ x ∣ - 12

.

(A)

\frac{128}{3}

(B)

\frac{176}{3}

(C)

\frac{256}{3}

(D)

108

(E)

144

(F)

288

Voir la correction

For

x \geq 0

, the curve is

y = x^{2} - 4 x - 12 = (x - 6) (x + 2)

, which crosses the

y = x^{2} - 4 x - 12 = (x - 6) (x + 2)

-axis at

x = 6

(and

x = - 2

, outside the region

x \geq 0

). For

x < 0

, the curve is

y = x^{2} + 4 x - 12 = (x + 6) (x - 2)

, which crosses the

y = x^{2} + 4 x - 12 = (x + 6) (x - 2)

-axis at

x = - 6

(and

x = 2

, outside

x < 0

). The finite enclosed area lies between

x = - 6

and

x = 6

. By symmetry about the

y

-axis, the total area is twice the area for

x \in [0, 6]

. For

0 \leq x \leq 6

, the curve is below the

x

-axis, so:

\int_{0}^{6} (x^{2} - 4 x - 12) d x = [\frac{x ^{3}}{3} - 2 x^{2} - 12 x]_{0}^{6} = \frac{216}{3} - 72 - 72 = 72 - 144 = - 72

. The unsigned area for

x \in [0, 6]

is

72

, so the total area is

2 \times 72 = 144

. Hence the answer is E.