Oxford MAT Livestream 2024

Integration & Differentiation – Solutions

Revision Questions
1. The derivative of xa is axa−1 so the derivative of this expression is 17x16 + 17x−18 .
√ √
2. Remember that x = x1/2 and 3 x = x1/3 , so the derivative of this expression is
1 1
x−1/2 + x−2/3 , which we could write as √ + 2/3 if we wanted to.
x x
3. Remember that the derivative of a constant is 0, so the derivative of this expression is
just −3e3x .
dy
4. We need to find the value of the derivative at x = 2 because that’s equal to the
dx
dy
gradient of the tangent. We can differentiate to find = ex + 2x so that gradient we
dx
wanted is e2 + 4. We also want the tangent to have the same value at x = 2 as the
curve; that’s ex + x2 at x = 2, which is also e2 + 4. So our tangent is y = (e2 + 4)(x − 1).

5. First find the derivative at x = 3, which is 6 for this parabola. That’s the gradient of
1
the tangent, and the normal is at right-angles to the tangent, so it has gradient − .
6
x
We have y = − + c and we want the normal to go through the point (3, 9). So we
6
19
want c = .
2
dy
6. The turning points must have = 0 so we must have 4x3 − 6x2 + 2x = 0. That
dx
happens when x = 0 or when 2x2 − 3x + 1 = 0 which happens when (2x − 1)(x − 1) = 0,
which is either x = 1 or x = 12 .
Now find the second derivative to check whether these are minima or maxima. We have
d2 y
= 12x2 − 12x + 2, which is positive for x = 0, negative for x = 12 , and positive
dx2
for x = 1. So we have a (local) minimum, then a (local) maximum, then a (local)
minimum.
The function is decreasing for x < 0, then increasing for 0 < x < 21 , then decreasing for
1
2
< x < 1 then increasing for x > 1.

7. The line definitely goes through A, which doesn’t move. The thing we learn from
“differentiation from first principles” is that the gradient of the line gets closer and
closer to the derivative of the function at A.
The derivative is 3x2 + 2x + 1 which is 6 at x = 1. The value is 4, so the tangent is
y = 6x − 2. So if the line through A and B is y = mx + c then m gets closer and closer
to 6 and c gets closer and closer to −2.

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8. First find the points where y = 0. We have (x + 3)(x + 1) = 0 so these points are at
x = −1 or x = −3. In between, we have y < 0 (by considering the graph).
Z −1
So we want − x2 +4x+3 dx. That minus sign out the front is because the function
−3
4
is negative in this region. This works out to be .
3
Z Z
x+3 1 3 1 3
9. • dx = + , dx = − − + c where c is a constant.
x3 x2 x3 x 2x2
√
Z Z
3
• 3
x dx = x1/3 dx = x4/3 + c where c is a constant.
4
x31
Z  Z

3 5

• x2 dx = x30 dx = + c where c is a constant.
31
x7 3x5
Z Z

3
• x + 1 dx = x6 + 3x4 + 3x2 + 1 dx =
2
+ + x3 + x + c where c is a
7 5
constant.
10. The graph of f (−x) is the graph of f (x) reflected in y-axis. Also, note that if we reflect
the interval −1 ⩽ x ⩽ 1 in the y-axis then we get the same interval back. On the
left-hand side, we’re finding the area under f (x) (or maybe negative the area in any
regions where f is negative). On the right-hand side, we’re calculating exactly the same
area, but with the shape of the graph reflected.
1
11. First consider the graph y = . The area under the graph between x = 1 and x = 10
x
is I1 . Now consider stretching that region by a factor of 10 parallel to the x-axis, and
squashing it by a factor of 10 parallel to the y-axis. The area will be the same, and
1
(amazingly!) any point that was on the curve y = is still on the graph after these
x
transformations. So I2 , the area under the graph between 10 and 100 is equal to I1 .
This means that
Z 100 Z 10 Z 100
1 1 1
dx = dx + dx = I1 + I2 = 2I1 .
1 x 1 x 10 x
But similarly,
Z 1000 if we think about stretching the graph again in the same way, we find
1
that dx is also equal to I1 . By setting N to be a large power of ten, we can
Z100N x
1
make dx arbitrarily large.
1 x
x2 1 R3
12. Note that + = 1 so I3 + I4 = 1
1 dx = 2. So I3 + I4 = 2.
1 + x2 1 + x2
Z 3
x4 2 x2 2
13. Note that 2
=x − 2
so this new integral is x2 dx − I4 = 8 − I4 .
1+x 1+x 1 3

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MAT Questions
MAT 2017 Q1A

• Stationary points at those values of x for which the derivative of f (x) is zero. So we’re
looking for points where 6x2 − 2kx + 2 = 0

• We would like to know whether or not there are two distinct values of x that satisfy
that equation. It’s a quadratic equation.

• We should check the discriminant. If (2k)2 − 4 × 6 × 2 > 0 then there are two distinct
real solutions.
2
√
• That inequality
√ simplifies to k > 12. This is true when either k > 12 or when
k < − 12.

• The answer is (b).

MAT 2018 Q1A

√
• The curve and the line have the same value where x = x − 2. We could square both
sides to find that x = (x − 2)2 , so x2 − 3x + 4 =√0. The solutions to that equation x = 4
and x = 1. But we should check our answers; 1 = 1 and 1 − 2 =√−1, so x = 1 is not
a solution. The point with x = 4 is a genuine solution, because 4 really is equal to
4 − 2.
√ √
• If we integrate x from 0 to 4, we would get the area between the curve x and the
x-axis, in the region 0 < x < 4. We want something slightly different, because we don’t
want the bit of that area which is under the line y = x − 2.

• That area under the line y = x − 2 is a right-angled triangle with base 2 and height 2,
so it has area 2. (We could integrate from 2 to 4 to get that area, but I know a formula
for the area of a triangle).
R4√
• So we just need 0 x dx − 2.

• Time for some integration;

4 4
√
Z 
2 16 10
x dx − 2 = x3/2 −2= −2= .
0 3 0 3 3

• The answer is (d).

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MAT 2018 Q1G

√
• We can write the second curve as y = ± x. The curve only exists where x > 0, and in
that region the first curve y = x2 + c has positive gradient. So√the gradient will only
match if we consider the part of the second curve where y = + x with a + sign.

• Now we can match up the values of the curves, and separately match up the gradients
of the curves, for the system of equations;
√ 1 1
x2 + c = x, 2x = √ ,
2 x

where by x, I mean the particular value of x at the point where the curves meet. (Really
I should give that value of x a name, I’m being lazy with my notation here)
−2/3
• I can solve the second equation
√ for x = 4
2 . Then substituting that value into the
first equation gives me c = 4−2/3 − 4−2/3 .

• This simplifies to c = 4−1/3 − 4−4/3 .

• Most of the options are a single expression, so I’m looking for a way to simplify further.
Eventually I spot that 43 = 1 + 31 , so the second term in my expression is 4−1 4−1/3

• So c = 3
4
× 4−1/3

• The answer is (b).

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MAT 2009 Q3

(i) The function inside the brackets is x3 − 1. That’s 0 at x = 1 and it’s −1 at x = 0. It’s
negative for x < 1. Here’s my sketch of the square of that function, on the left below.

y y

4 4

2 2

x x
−1 1 −1 1

(ii) For higher powers of n, the function x2n−1 is approximately zero for |x| < 1, and it
grows rapidly outside that range. So x2n−1 − 1 is about −1 for the range |x| < 1, but it
shoots up to high positive values soon after x = 1 and it shoots down to very negative
values just before x = −1. If we square that function, we’ll get something that’s about
1 for most of the range |x| < 1, but near the edges of that region two strange things
will happen. Near x = −1, the function inside the brackets just gets really negative.
For x near 1, the function inside the brackets increases to zero then increases to high
positive values. For the square of the function, this is a decrease to zero first, then an
increase to high positive values. See my sketch above, on the right, with the dashed
line indicating my previous sketch.

(iii) We have
1 1 1
x4n−1 x2n
Z Z 
4n−2 2n−1 1 1
fn (x) dx = x − 2x + 1 dx = − +x = − +1
0 0 4n − 1 n 0 4n − 1 n

where the contributions from the lower limit x = 0 are all zero because those powers of
x give zero for n ⩾ 1 a whole number.

(iv) We would like

1 1 A
− ⩽1− 1+
4n − 1 n n+B
for all n ⩾ 1. This rearranges (being careful not to multiply by negative numbers) to
the inequality
0 ⩾ (4A − 3)n2 + (1 − A − 3B)n + B
If the coefficient of n2 is positive, this clearly doesn’t work (because the right-hand side
will get really large and positive for large enough n), so we must have A ⩽ 3/4.

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(v) If the coefficient of n2 is zero, then A = 3/4. We still need

 
1
0⩾ − 3B n + B
4

for all n ⩾ 1. For this to work, the linear function on the right-hand side must have
negative or zero gradient, and the value at n = 1 must be negative or zero. We therefore
require both B ⩾ 1/12 and also B ⩾ 1/8. Since we need both of these to hold, we must
have B ⩾ 1/8.
Quick check that if A = 3/4 and B = 1/8 then the inequality is in fact true.

Extension

• If n = 1/2 then the function is constant and zero. The integral is zero.

• We need n > 1/4. If n ⩽ 1/4 then the integral doesn’t exist, because x4n−1 gets very
large near x = 0.

• Sketch:

y
2

x
1 2 3 4

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MAT 2015 Q3
(i) There are lots of choices that work! To keep things simple, I picked constant functions
1 1 1
f (x) = 200 and g(x) = 0. Then |f (x) − g(x)| is less than 100 , but not less than 320 .
(ii) In this case, f (x) − g(x) = sin(4x2 )/400. The 4x2 inside the brackets doesn’t really
matter; whatever the value of θ, we have | sin(θ)| < 1. So in this case we have
|f (x) − g(x)| ⩽ 1/400, which is less than 1/320 for all x.
x2 x3 x4
(iii) I can integrate for g(x) = 1 + x + + + .
2 6 24
The value of |g(x) − f (x)| is x4 /24, but that can only be as large as (1/2)4 /24, and
that’s less than 1/320.
(iv) I’d like something involving h(x) and g(x), so that they appear on opposite sides of the
equation. I’ll take the difference between the defining equations for h(x) and g(x) for
Z x Z x
h(x) − g(x) = 1 − 1 + h(t) dt − f (t) dt.
0 0

This rearranges to the target expression, if I subtract f (x) from both sides.
(v) There’s a given range for x. The largest value of the integral would come if, hypotheti-
cally, x = 21 and if h(t) − f (t) were equal to its maximum value of h(x0 ) − f (x0 ) all the
way from t = 0 to t = 21 . That would give an area of 12 × (h(x0 ) − f (x0 )).
(vi) We’ve been told that h(x) ⩾ f (x), so h(x) − f (x) ⩾ 0. We just need to check that
h(x) − f (x) ⩽ 1/100. It’s enough to check that the maximum value h(x0 ) − f (x0 ) is
less than 1/100.
In part (v) we worked out a fact about the integral in part (iv). Together, we have
1
h(x) − f (x) ⩽ g(x) − f (x) + (h(x0 ) − f (x0 )) .
2
But if we set x = x0 and rearrange, this shows that 12 (h(x0 ) − f (x0 )) ⩽ g(x0 ) − f (x0 ).
And we know from part (iii) that the last expression there is less than 1/320. So the
expression h(x0 ) − f (x0 ) is less than 1/100 and we have a good approximation.

Extension
Rx x
• If h(t) = et then the right-hand side is 1 + 0
et dt = 1 + [et ]0 = ex .
• Perhaps we could try h(x) = Aekx . Then the right-hand side becomes
Z x x
3Aekt

kt 3A kx

2+ 3Ae dt = 2 + =2+ e −1 .
0 k 0 k
If k = 3 and A = 2 then this simplifies to 2e3x which is h(x).
So h(x) = 2e3x is a solution.

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