Exam answer Show all explanations
(a)
Let I = ∫ x 2 ( x 2 + 4 ) ( x 2 + 9 ) d x I = \displaystyle\int\frac{x^2}{(x^2 + 4)(x^2 + 9)}\,dx I = ∫ ( x 2 + 4 ) ( x 2 + 9 ) x 2 d x
Put x 2 = t x^2 = t x 2 = t
t ( t + 4 ) ( t + 9 ) = A t + 4 + B t + 9 \frac{t}{(t + 4)(t + 9)} = \frac{A}{t + 4} + \frac{B}{t + 9} ( t + 4 ) ( t + 9 ) t = t + 4 A + t + 9 B Why?
The integrand contains x x x x 2 x^2 x 2 x 2 x^2 x 2 t t t x x x d x dx d x
In t t t
⇒ t = A ( t + 9 ) + B ( t + 4 ) \Rightarrow t = A(t + 9) + B(t + 4) ⇒ t = A ( t + 9 ) + B ( t + 4 ) Why?
Multiply both sides by ( t + 4 ) ( t + 9 ) (t + 4)(t + 9) ( t + 4 ) ( t + 9 )
Putting t = − 4 t = -4 t = − 4 − 4 = 5 A -4 = 5A − 4 = 5 A A = − 4 5 A = -\frac45 A = − 5 4 t = − 9 t = -9 t = − 9 − 9 = − 5 B -9 = -5B − 9 = − 5 B B = 9 5 B = \frac95 B = 5 9
Why?
Choose values of t t t t = − 4 t = -4 t = − 4 t + 9 = 5 t + 9 = 5 t + 9 = 5 t = − 9 t = -9 t = − 9 t + 4 = − 5 t + 4 = -5 t + 4 = − 5
At t = − 4 t = -4 t = − 4
− 4 = A ( − 4 + 9 ) + B ( − 4 + 4 ) − 4 = A ( 5 ) + B ( 0 ) − 4 = 5 A A = − 4 5 = − 4 5 \begin{aligned} -4 &= A(-4 + 9) + B(-4 + 4) \\ -4 &= A(5) + B(0) \\ -4 &= 5A \\ A &= \frac{-4}{5} = -\frac45 \end{aligned} − 4 − 4 − 4 A = A ( − 4 + 9 ) + B ( − 4 + 4 ) = A ( 5 ) + B ( 0 ) = 5 A = 5 − 4 = − 5 4 At t = − 9 t = -9 t = − 9
− 9 = A ( − 9 + 9 ) + B ( − 9 + 4 ) − 9 = A ( 0 ) + B ( − 5 ) − 9 = − 5 B B = − 9 − 5 = 9 5 \begin{aligned} -9 &= A(-9 + 9) + B(-9 + 4) \\ -9 &= A(0) + B(-5) \\ -9 &= -5B \\ B &= \frac{-9}{-5} = \frac95 \end{aligned} − 9 − 9 − 9 B = A ( − 9 + 9 ) + B ( − 9 + 4 ) = A ( 0 ) + B ( − 5 ) = − 5 B = − 5 − 9 = 5 9 I = − 4 5 ∫ d x x 2 + 2 2 + 9 5 ∫ d x x 2 + 3 2 \begin{aligned} I &= -\frac45\int\frac{dx}{x^2 + 2^2} \\ &\quad + \frac95\int\frac{dx}{x^2 + 3^2} \end{aligned} I = − 5 4 ∫ x 2 + 2 2 d x + 5 9 ∫ x 2 + 3 2 d x Why?
Put back t = x 2 t = x^2 t = x 2 t + 4 = x 2 + 2 2 t + 4 = x^2 + 2^2 t + 4 = x 2 + 2 2 t + 9 = x 2 + 3 2 t + 9 = x^2 + 3^2 t + 9 = x 2 + 3 2
x 2 ( x 2 + 4 ) ( x 2 + 9 ) = − 4 5 x 2 + 4 + 9 5 x 2 + 9 \begin{aligned} &\frac{x^2}{(x^2 + 4)(x^2 + 9)} \\ &= \frac{-\frac45}{x^2 + 4} + \frac{\frac95}{x^2 + 9} \end{aligned} ( x 2 + 4 ) ( x 2 + 9 ) x 2 = x 2 + 4 − 5 4 + x 2 + 9 5 9 The constants − 4 5 -\frac45 − 5 4 9 5 \frac95 5 9
Using ∫ d x x 2 + a 2 = 1 a tan − 1 x a + C \int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + C ∫ x 2 + a 2 d x = a 1 tan − 1 a x + C
I = − 4 5 ⋅ 1 2 tan − 1 x 2 + 9 5 ⋅ 1 3 tan − 1 x 3 + C \begin{aligned} I &= -\frac45\cdot\frac12\tan^{-1}\frac{x}{2} \\ &\quad + \frac95\cdot\frac13\tan^{-1}\frac{x}{3} + C \end{aligned} I = − 5 4 ⋅ 2 1 tan − 1 2 x + 5 9 ⋅ 3 1 tan − 1 3 x + C Why?
The first integral has a = 2 a = 2 a = 2 a = 3 a = 3 a = 3
4 5 ⋅ 1 2 = 2 5 , 9 5 ⋅ 1 3 = 3 5 \frac45\cdot\frac12 = \frac25,\quad \frac95\cdot\frac13 = \frac35 5 4 ⋅ 2 1 = 5 2 , 5 9 ⋅ 3 1 = 5 3 Top times top, bottom times bottom, then cancel:
4 5 ⋅ 1 2 = 4 × 1 5 × 2 = 4 10 = 4 ÷ 2 10 ÷ 2 = 2 5 9 5 ⋅ 1 3 = 9 × 1 5 × 3 = 9 15 = 9 ÷ 3 15 ÷ 3 = 3 5 \begin{aligned} \frac45\cdot\frac12 &= \frac{4 \times 1}{5 \times 2} = \frac{4}{10} \\ &= \frac{4 \div 2}{10 \div 2} = \frac25 \\[4pt] \frac95\cdot\frac13 &= \frac{9 \times 1}{5 \times 3} = \frac{9}{15} \\ &= \frac{9 \div 3}{15 \div 3} = \frac35 \end{aligned} 5 4 ⋅ 2 1 5 9 ⋅ 3 1 = 5 × 2 4 × 1 = 10 4 = 10 ÷ 2 4 ÷ 2 = 5 2 = 5 × 3 9 × 1 = 15 9 = 15 ÷ 3 9 ÷ 3 = 5 3 The first constant keeps its minus sign: − 2 5 -\frac25 − 5 2
∴ I = − 2 5 tan − 1 x 2 + 3 5 tan − 1 x 3 + C \therefore\ I = -\frac25\tan^{-1}\frac{x}{2} + \frac35\tan^{-1}\frac{x}{3} + C ∴ I = − 5 2 tan − 1 2 x + 5 3 tan − 1 3 x + C
(b)
Let I I I
For 1 ≤ x ≤ 3 1 \le x \le 3 1 ≤ x ≤ 3 ∣ x − 1 ∣ = x − 1 |x - 1| = x - 1 ∣ x − 1∣ = x − 1 ∣ x − 3 ∣ = − ( x − 3 ) |x - 3| = -(x - 3) ∣ x − 3∣ = − ( x − 3 ) ∣ x − 2 ∣ = − ( x − 2 ) |x - 2| = -(x - 2) ∣ x − 2∣ = − ( x − 2 ) [ 1 , 2 ] [1, 2] [ 1 , 2 ] ∣ x − 2 ∣ = x − 2 |x - 2| = x - 2 ∣ x − 2∣ = x − 2 [ 2 , 3 ] [2, 3] [ 2 , 3 ]
Why?
To integrate a modulus, first remove it: ∣ u ∣ = u |u| = u ∣ u ∣ = u u ≥ 0 u \ge 0 u ≥ 0 ∣ u ∣ = − u |u| = -u ∣ u ∣ = − u u < 0 u < 0 u < 0
On [ 1 , 3 ] [1, 3] [ 1 , 3 ] x − 1 ≥ 0 x - 1 \ge 0 x − 1 ≥ 0 x − 3 ≤ 0 x - 3 \le 0 x − 3 ≤ 0 x − 2 x - 2 x − 2 x = 2 x = 2 x = 2
I = ∫ 1 3 ( x − 1 ) d x + ∫ 1 2 − ( x − 2 ) d x + ∫ 2 3 ( x − 2 ) d x − ∫ 1 3 ( x − 3 ) d x \begin{aligned} I &= \int_1^3(x - 1)\,dx \\ &\quad + \int_1^2-(x - 2)\,dx \\ &\quad + \int_2^3(x - 2)\,dx \\ &\quad - \int_1^3(x - 3)\,dx \end{aligned} I = ∫ 1 3 ( x − 1 ) d x + ∫ 1 2 − ( x − 2 ) d x + ∫ 2 3 ( x − 2 ) d x − ∫ 1 3 ( x − 3 ) d x Why?
The middle term is split using the property
∫ a b f d x = ∫ a c f d x + ∫ c b f d x \int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx ∫ a b f d x = ∫ a c f d x + ∫ c b f d x with a = 1 a = 1 a = 1 c = 2 c = 2 c = 2 b = 3 b = 3 b = 3
I = ∫ 1 3 2 d x + ∫ 1 2 ( 2 − x ) d x + ∫ 2 3 ( x − 2 ) d x \begin{aligned} I &= \int_1^3 2\,dx + \int_1^2(2 - x)\,dx \\ &\quad + \int_2^3(x - 2)\,dx \end{aligned} I = ∫ 1 3 2 d x + ∫ 1 2 ( 2 − x ) d x + ∫ 2 3 ( x − 2 ) d x Why?
The first and last integrals have the same limits, so their integrands combine: ( x − 1 ) − ( x − 3 ) = 2 (x - 1) - (x - 3) = 2 ( x − 1 ) − ( x − 3 ) = 2 − ( x − 2 ) = 2 − x -(x - 2) = 2 - x − ( x − 2 ) = 2 − x
( x − 1 ) − ( x − 3 ) = x − 1 − x + 3 = ( x − x ) + ( − 1 + 3 ) = 0 + 2 = 2 − ( x − 2 ) = − x + 2 = 2 − x \begin{aligned} &(x - 1) - (x - 3) \\ &= x - 1 - x + 3 \\ &= (x - x) + (-1 + 3) \\ &= 0 + 2 = 2 \\[4pt] &-(x - 2) = -x + 2 = 2 - x \end{aligned} ( x − 1 ) − ( x − 3 ) = x − 1 − x + 3 = ( x − x ) + ( − 1 + 3 ) = 0 + 2 = 2 − ( x − 2 ) = − x + 2 = 2 − x I = [ 2 x ] 1 3 + [ ( 2 − x ) 2 − 2 ] 1 2 + [ ( x − 2 ) 2 2 ] 2 3 = ( 6 − 2 ) + ( 0 + 1 2 ) + ( 1 2 − 0 ) \begin{aligned} I &= \Big[2x\Big]_1^3 + \left[\frac{(2 - x)^2}{-2}\right]_1^2 \\ &\quad + \left[\frac{(x - 2)^2}{2}\right]_2^3 \\ &= (6 - 2) + \left(0 + \frac12\right) \\ &\quad + \left(\frac12 - 0\right) \end{aligned} I = [ 2 x ] 1 3 + [ − 2 ( 2 − x ) 2 ] 1 2 + [ 2 ( x − 2 ) 2 ] 2 3 = ( 6 − 2 ) + ( 0 + 2 1 ) + ( 2 1 − 0 ) Why?
For a linear expression inside a power,
∫ ( p x + q ) d x = ( p x + q ) 2 2 p \int(px + q)\,dx = \frac{(px + q)^2}{2p} ∫ ( p x + q ) d x = 2 p ( p x + q ) 2 (plus a constant). For 2 − x 2 - x 2 − x p = − 1 p = -1 p = − 1 − 2 -2 − 2
Watch the signs: at x = 1 x = 1 x = 1 ( 2 − 1 ) 2 − 2 = − 1 2 \frac{(2 - 1)^2}{-2} = -\frac12 − 2 ( 2 − 1 ) 2 = − 2 1 0 + 1 2 0 + \frac12 0 + 2 1
Each bracket is (value at the top limit) minus (value at the bottom limit):
[ 2 x ] 1 3 = 2 ( 3 ) − 2 ( 1 ) = 6 − 2 = 4 [ ( 2 − x ) 2 − 2 ] 1 2 = ( 2 − 2 ) 2 − 2 − ( 2 − 1 ) 2 − 2 = 0 − 2 − 1 − 2 = 0 − ( − 1 2 ) = 0 + 1 2 [ ( x − 2 ) 2 2 ] 2 3 = ( 3 − 2 ) 2 2 − ( 2 − 2 ) 2 2 = 1 2 − 0 = 1 2 \begin{aligned} \Big[2x\Big]_1^3 &= 2(3) - 2(1) \\ &= 6 - 2 = 4 \\[4pt] \left[\frac{(2 - x)^2}{-2}\right]_1^2 &= \frac{(2 - 2)^2}{-2} - \frac{(2 - 1)^2}{-2} \\ &= \frac{0}{-2} - \frac{1}{-2} \\ &= 0 - \Big(-\frac12\Big) = 0 + \frac12 \\[4pt] \left[\frac{(x - 2)^2}{2}\right]_2^3 &= \frac{(3 - 2)^2}{2} - \frac{(2 - 2)^2}{2} \\ &= \frac12 - 0 = \frac12 \end{aligned} [ 2 x ] 1 3 [ − 2 ( 2 − x ) 2 ] 1 2 [ 2 ( x − 2 ) 2 ] 2 3 = 2 ( 3 ) − 2 ( 1 ) = 6 − 2 = 4 = − 2 ( 2 − 2 ) 2 − − 2 ( 2 − 1 ) 2 = − 2 0 − − 2 1 = 0 − ( − 2 1 ) = 0 + 2 1 = 2 ( 3 − 2 ) 2 − 2 ( 2 − 2 ) 2 = 2 1 − 0 = 2 1 Add the three parts:
4 + 1 2 + 1 2 = 4 + 1 = 5 \begin{aligned} 4 + \frac12 + \frac12 &= 4 + 1 = 5 \end{aligned} 4 + 2 1 + 2 1 = 4 + 1 = 5 ∴ I = 4 + 1 2 + 1 2 = 5 \therefore\ I = 4 + \dfrac12 + \dfrac12 = 5 ∴ I = 4 + 2 1 + 2 1 = 5
Where this comes from Chapter 7, Integrals : Partial fractions; definite integral of |x − a|.
Question type: (a) Partial fractions ; (b) A modulus inside the integral
How to spot it: click to see (a) Ch 7, type 7. Partial fractions: break a fraction into simple pieces
Spot it: a fraction of two polynomials whose bottom splits into factors.
Here: The bottom ( x 2 + 4 ) ( x 2 + 9 ) (x^2 + 4)(x^2 + 9) ( x 2 + 4 ) ( x 2 + 9 ) x x x x 2 x^2 x 2 t t t x 2 x^2 x 2 A = − 4 5 A = -\frac45 A = − 5 4 B = 9 5 B = \frac95 B = 5 9 tan − 1 \tan^{-1} tan − 1
(b) Ch 8, type 6. A modulus inside the integral: split at the corner
Spot it: bars inside the integral, like ∣ x + 3 ∣ |x + 3| ∣ x + 3∣ x ∣ x ∣ x|x| x ∣ x ∣
Here: On [ 1 , 3 ] [1, 3] [ 1 , 3 ] ∣ x − 1 ∣ = x − 1 |x - 1| = x - 1 ∣ x − 1∣ = x − 1 ∣ x − 3 ∣ = 3 − x |x - 3| = 3 - x ∣ x − 3∣ = 3 − x x − 2 x - 2 x − 2 x = 2 x = 2 x = 2
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