Mathematics · Oct / Nov 2021 · Paper 11 · Question 1

∫

1

3

x

4

1

3

x

−

3

1

12

x

3

+

−

d

x

=

12

×

+

×

2

x

−

× −

2

4

6

2

3

6

x

2

x

4

1

1

1

3

4

−

3

=

x

+

x

2

+

x

+

c

3

2

∫

x

n

d

x

→

x

n

+

1

M1

Applies

for at least one index.

1

1

1

3

4

2

−

The index must be processed so allow for

x

→

x

or

→

x

or

→

x

x

4

x

1

1

1

A1

One correct term simplified or + c. Look for one of

3

x

4

,

+

x

,

+

x

−

3

or the + c.

2

3

2

A1

Two correct terms simplified or one correct simplified with + c.

1

1

1

Look for two of

3

x

4

,

+

3 x

,

+

2 x

−

3

,

+

c

2

A1

Three correct terms simplified or two correct simplified with + c.

1

1

1

Look for three of

3

x

4

,

+

3 x

,

+

2 x

−

3

,

+

c

2

1

1

1

A1

3

x

4

+

x

+

x

−

3

+

c

all correct and simplified and on one line.

2

3

2

1

1

1

1

1

Allow simplified equivalents such as

x

↔

x

and

x

−↔

3

2

3

3

2

3

2

x

Award once a correct expression is seen and isw but if there is any additional/incorrect

notation and no correct expression has been seen on its own, withhold the final mark.

∫

1

1

1

1

1

1

E.g.

3

x

4

+

x

+

x

−

3

+

c x

d , 3

x

4

+

x

+

x

−

3

+

c

=

0

2

2

3

2

3

2