Understanding Engineering Mathematics

For example, because we know that the derivative ofx^2 is 2x, so we know that the
integral of 2xisx^2. Actually, because the derivative of a constant is zero the most general
integral of 2xisx^2 +CwhereCis anarbitrary constant of integration.We write

∫

2 xdx=x^2 +C

Solution to review question 9.1.1

A.Hopefully, this won’t give you much trouble now:

(i)

d

dx

(x^3 )= 3 x^2

(ii)

d

dx

(sin( 2 x))=2cos( 2 x)

by the function of a function rule.

(iii)

d

dx

( 2 e^3 x)= 2 e^3 x× 3 = 6 e^3 x

B. Of course, these questions are not unrelated toA!

(i) Having just done it inA(i) we know that we can obtain 3x^2 by

differentiatingx^3 :

d

dx

(x^3 )= 3 x^2

And in fact we get the same result if we add an arbitrary constant

Ctox^3 :

d

dx

(x^3 +C)= 3 x^2

It follows that the most general integral of 3x^2 is:

∫

3 x^2 dx=x^3 +C

(ii) A little more thought is needed to integrate cos 2x. Differentiating

sin 2xgives us 2 cos 2xnotcos 2x. To get the latter we should

differentiate^12 sin 2x

d

dx

( 1

2 sin 2x

)

=

1

2

2cos2x=cos 2x

Remembering the arbitrary constant we thus have:

∫

cos 2xdx=

1

2

sin 2x+C

(iii) Again, we have to fiddle with the multiplier of thee^3 x.Since

d

dx

(e^3 x)= 3 e^3 x