Integration

8 methods · 9 questions

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Before integrating a polynomial over symmetric limits $[- a, a]$ , check the parity of the terms. Even powers integrate to twice the integral over $[0, a]$ , and odd powers integrate to $0$ and disappear entirely, cutting the computation in half and avoiding sign errors at the lower limit.

Trigger: a definite integral with symmetric limits $\int_{- a}^{a}$ , often after a coordinate shift.

Instances:
(i) for an even integrand $f (x) = f (- x)$ , replace $\int_{- a}^{a} f (x) d x$ with $2 \int_{0}^{a} f (x) d x$ ;
(ii) for an odd term like $x$ or $x^{3}$ , cross it out before integrating;
(iii) constant denominators can be factored out before exploiting the symmetry.

Linked questions (2)

NSAA 2018 Section 1, Part E (Mathematics questions) · Q73

Curve

C

has equation

y = 9 - x^{2}

Line

L

has equation

y = 5

What is the area enclosed between

C

and

L

A $\frac{32}{3}$
B $\frac{62}{3}$
C $\frac{92}{3}$
D $\frac{122}{3}$
E $\frac{152}{3}$

Worked solution

Answer:A

The curve and the line meet where

9 - x^{2} = 5

, that is

x^{2} = 4

, so at

x = - 2

and

x = 2

.

Between those values

9 - x^{2} > 5

, so the curve lies above the line and the enclosed area is

\int_{- 2}^{2} [(9 - x^{2}) - 5] d x = \int_{- 2}^{2} (4 - x^{2}) d x .

The integrand is even, so the integral is twice the integral over

[0, 2]

2 [4 x - \frac{x ^{3}}{3}]_{0}^{2} = 2 (8 - \frac{8}{3}) = 2 \times \frac{16}{3} = \frac{32}{3} .

The answer is A.

ENGAA 2023 Section 1, Part B (Mathematics questions) · Q31

Which of the following is equal to

\int_{- (3 - 5)}^{3 - 5} \frac{x ^{2}}{3 - 5} d x

A $0$
B $\frac{2}{3}$
C $\frac{8}{3}$
D $4$
E $\frac{28 - 12 5}{3}$
F $28 - 125$

Worked solution

Answer:E

Write

k = 3 - 5

. The denominator is a constant, so it comes out of the integral, and the limits are

- k

and

k

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

Now

k^{2} = (3 - 5)^{2} = 9 - 65 + 5 = 14 - 65,

so the integral equals

\frac{2 ( 14 - 6 5 )}{3} = \frac{28 - 12 5}{3} .

As a check, the integrand is even and strictly positive away from

0

, so the value must be positive: with

k \approx 0.764

the integral is about

0.389

, and

\frac{28 - 12 5}{3} \approx \frac{1.167}{3} \approx 0.389

.

The answer is E.