ABSTRACT
Problem 2. ∫
When a = 2 and n = 3,∫ 2t3 dt = 2t
3+ 1 + c = 2t4
4 + c = 1
2 t4 + c
Note that each of the results in worked examples 1 and 2 may be checked by differentiating them.
Problem 3. Determine ∫
8dx
∫ 8dx is the same as
∫ 8x0 dx and, using the general
rule when a = 8 and n = 0, gives∫ 8x0 dx = 8x
0+ 1 + c = 8x + c
In general, if k is a constant then ∫ kdx = kx + c
Problem 4. Determine ∫
2x dx
When a = 2 and n = 1,∫ 2x dx =
∫ 2x1 dx = 2x
1+ 1 + c = 2x2
2 + c
= x2 + c
Problem 5. Determine ∫ (
3+ 2 5 x − 6x2
) dx
∫ ( 3+ 2
5 x − 6x2
) dx may be written as
∫ 3dx+ ∫ 25 x dx − ∫ 6x2 dx
i.e. each term is integrated separately. (This splitting up of terms only applies, however, for addition and subtraction.) Hence,∫ (
3+ 2 5 x − 6x2
) dx
= 3x + ( 2 5
) x1+1
1+ 1 − (6) x2+1
2+ 1 + c
= 3x + ( 2 5
) x2
2 − (6) x
3 + c = 3x + 1
5 x2 − 2x3 + c
Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant c at the end is sufficient.
