# 1. Anti-Derivatives

If ƒ = dF/dx , we call F the anti-derivative (or indefinite integral) of ƒ.

Example 1: If ƒ (x) = x, we can find its anti-derivative by realising that for F (x) = 1/2 x2:

ƒ = dF/dx = d/dx (1/2 x2) = 1/2 × 2x = x = ƒ (x)

Thus, F (x) = 1/2 x2 is an anti-derivative of ƒ (x) = x.

However, if C is a constant:

d/dx (1/2 x2 + C) = 1/2 × 2x = x

since the derivative of a constant is zero. The general anti-derivative of x is thus 1/2 x2 + C, where C can be any constant.

Note that you should always check an anti-derivative F by differentiating it and seeing that you recover ƒ.

Quiz 1: Using d(xn)/dx = nxn−1, select an anti-derivative of x6:

(a)6x5 Incorrect - please try again!
(b) 1/5x5 Incorrect - please try again!
(c) 1/7x7 Correct - well done!
(d) 1/6x7 Incorrect - please try again!

Explanation: To find the anti-derivative of x6 first calculate the derivative of F (x) = 1/7 x7. Using the basic formula:

d(xn)/dx = nxn−1

with n = 7, we have:

 dF/dx = d/dx(1/7 x7 ) = 1/7 d/dx ( x7 ) = 1/7 × 7x7−1 = x6

This result shows that the function F (x) = 1/7x7 + C is the general anti-derivative of x6.

In general the anti-derivative or integral of xn is given by:

 if F (x) = xn , then F (x) = 1/n + 1 xn+1 for n ≠ −1

Note: This rule does not apply to 1 ⁄ x = x−1. Since the derivative of ln (x) is 1 ⁄ x, the anti-derivative of 1 ⁄ x is ln (x) - see later.

Also note that since 1 = x0, the rule says that the anti-derivative of 1 is x. This is correct since the derivative of x is 1.

We now introduce two important properties of integrals, which follow from the corresponding rules for derivatives.

If a is any constant and F (x) is the anti-derivative ƒ (x), then, from the module on Introductory Differentiation:

d/dx(aF (x)) = ad/dxF (x) = aƒ (x)

Thus:

 aF (x) is the anti-derivative of aƒ (x)

Quiz 2: Use this property to select the general anti-derivative of 3x1 ⁄ 2 from the choices below:

(a)2x3 ⁄ 2 + C Correct - well done!
(b) 3/2 x−1 ⁄ 2 + C Incorrect - please try again!
(c) 9/2 x3 ⁄ 2 + C Incorrect - please try again!
(d)6 x + C Incorrect - please try again!

Explanation: To find the general anti-derivative of 3x1 ⁄ 2, recall that for constant a the anti-derivative of aƒ (x) is aF (x), where F (x) is the anti-derivative of ƒ (x).

Thus, the anti-derivative of 3x½ is 3 × (the anti-derivative of x½).

To calculate the anti-derivative of x½ we recall that the anti-derivative of ƒ (x) = xn is F (x) = 1/ n + 1xn+1 for n ≠ 1. In our case n = ½ and therefore this result can be used. The anti-derivative of x½ is thus:

1/½ + 1 x(½+1) = 1/3 ⁄ 2 x3 ⁄ 2 = 1 × 2/3 x3 ⁄ 2 = 2/3 x3 ⁄ 2

Thus, the the general anti-derivative of 3x1 ⁄ 2 is 3 × 2/3 x3 ⁄ 2 + C = 2x3 ⁄ 2 +C

This result may be checked by differentiating F (x) = 2x3 ⁄ 2 +C

If dF/dx = ƒ (x) and dG/dx = g (x), then, from the sum rule of differentiation (see module on XXX):

d/dx(F + G) = d/dxF + d/dxG = ƒ (x) + g (x)

This leads to the sum rule for integration:

If F (x) is the anti-derivative of ƒ (x) and g (x) is the anti-derivative of g (x),
then F (x) + g (x) is the anti-derivative of ƒ (x) + g (x)

Note: Only one arbitrary constant C is needed in the anti-derivative of the sum of two (or more) functions.

Quiz 3: By using this property, which of the following is the general anti-derivative of 3x2 − 2x3?

(b) x31/2 x4 + CCorrect - well done!
(c) 3/2 x32/3 x4 + CIncorrect - please try again!
(d)x3 + 2/3 x + CIncorrect - please try again!

Explanation: To find the general anti-derivative of 3x2 − 2x3, we use the sum rule for anti-derivatives. The anti-derivative of 3x2 − 2x3 is (anti-derivative of 3x2) − (anti-derivative of 2x3). Since the anti-derivative of ƒ (x) = xn is F (x) = 1/n + 1 for n ≠ −1, forn = 2:

anti-derivative of x2 = 1/2 + 1 x2+1 = 1/3 x3

and so the anti-derivative of 3x2 is:

3 × (anti-derivative of x2) = 3 × 1/3 x3 = x3

Similarly, the anti-derivative of 2x3 is:

2 × (anti-derivative of x3) = = 2 × 1/3 + 1x3+1 = 1/2x4

Putting these results together we find that the general anti-derivative of 3x2 − 2x3 is:

F (x) = x31/3x4 + C

- which be confirmed by differentiation.

# 2. Indefinite Integral Notation

The notation for an anti-derivative or indefinite integral is:

if   F (x) = ƒ (x),   then  F (x) = ƒ (xdx + C

Here, is called the integral sign, while dx is called the measure and C is called the integration constant. We read this as "the integral of ƒ of x with respect to x" or "the integral of ƒ of x dx".

In other words, ∫ ƒ(x) dx means the general anti-derivative of ƒ(x) including an integration constant.

Example 2: To calculate the integral     x4dx , we recall that the anti-derivatve of xn for x ≠ −1 is xn + 1 ⁄ (n + 1). Here, n = 4, so we have:

x4dx = x4+1/4 + 1 + C = x5/5 + C

Quiz 4: Select the correct result for the indefinite integral   1/x  dx

(a)1/2 x− 3 ⁄ 2 + C Incorrect - please try again!
(b)2 x + C Correct - well done!
(c)1/2 x1 ⁄ 2 + C Incorrect - please try again!
(d) 2/x2 + C Incorrect - please try again!

Explanation: To calculate the indefinite integral

1/ x dx = 1/ x1 ⁄ 2 dx = x−1 ⁄ 2 dx

we recall the basic result, that the anti-derivative of ƒ (x) = xn is F (x) = 1/n + 1 xn+1 for n ≠ −1. In this case n = −½ and so:

 ∫ x−1 ⁄ 2dx = 1/− ½ + 1 x− ½ + 1 + C = 1/½ x½ + C = 1 × 2/1 x½ + C = 2 x½ + C = 2 √x + C

- where we recall that dividing by a fraction is equivalent to multiplying by its inverse (see the module on Fractions).

The previous rules for anti-derivatives may be expressed in integral notation as follows.

The integral of a function multiplied by any constant a is:

 ∫ aƒ (x) dx = a ∫ ƒ (x) dx

The sum rule for integration states that:

 ∫ (ƒ (x) + g (x)) dx =  ∫ ƒ (x) dx +  ∫ g (x) dx

To be able to integrate a greater number of functions, it is convenient first to recall the derivatives of some simple functions:

 y(x) sin (ax) cos (ax) eax ln (ax) dy/dx a cos (ax) −a sin (ax) a eax 1/x

Exercise 1: From the above table of derivatives calculate the indefinite integrals of the following functions:

(a) sin (ax)

Solution: From the table we have:

d/dx (sin (ax)) = acos (ax),  so:   d/dx ( − 1/a sin (ax) ) = cos (ax)

This implies:

cos (ax) = 1/a sin (ax) + C
(b)cos (ax)

Solution: From the table we can see that if y = cos (ax), then its derivative with respect to x is:

d/dx (cos (ax)) = −asin (ax),  so:   d/dx ( − 1/a cos (ax) ) = sin (ax)

Thus, we can conclude:

sin (ax) = − 1/a cos (ax) + C
(c) eax

Solution: From the table of derivatives:

d/dx (eax) = aeax,  so:   d/dx ( 1/a eax ) = eax

Thus the indefinite integral of eax is:

eax = 1/a eax + C
(d)1/x

Solution: From the table of derivatives see that the derivative of ln (x) with respect to x is:

d/dx (ln (x)) = 1/x

This implies that:

1/x = ln (x) + C

Click on questions to reveal their solutions

These results give the following table of indefinite integrals (the integration constants are omitted):

 y(x) xn (n ≠ −1) sin (ax) cos (ax) eax 1/x ∫ y(x) dx 1/n + 1 xn+1 − 1/a cos (ax) 1/a sin (ax) 1/a eax ln (x)

Exercise 2: From the above table, calculate the following integrals: DOING THIS ONE!

(a)    x7dx

Solution: We want to calculate    x7dx. From the table of indefinite integrals, for any n ≠ −1,

xn = 1/n + 1 xn+1

In the case of n = 7:

x7 = 1/7 + 1 × x7 + 1 + C = 1/8 x8 + C

Checking this:

d/dx ( 1/8 x8 + C ) = 1/8 d/dx x8 = 1/8 × 8 x7 = x7
(b)  2 sin (3x) dx

Solution: To calculate the integral   2 sin (3x) dx we use the formula:

sin (axdx = −  1/acos (ax)

In our case a = 3. Thus we have:

 ∫  2 sin (3x) dx = 2  ∫  sin (3x) dx = 2 ×  ( 1/3 )  cos (3x) + C = −  2/3 cos (3x) + C

Checking:

d/dx (− 2/3 cos (3x) + C) = − 2/3 d/dxcos (3x) = − 2/3  ×  (−3 sin (3x)) = 2 sin (3x)
(c)  4 cos (2x) dx

Solution: To calculate the integral   4 cos (2x) dx we use the formula:

cos (axdx = 1/a sin (ax)

with a = 2. This yields:

 ∫  4 cos (2x) dx = 4  ∫  cos (2x) dx = 4 ×  ( 1/2 )  sin (2x) + C

It may be checked that:

d/dx (2 sin (2x) + C) = 2 d/dxsin (2x) = 2 × ( (2 cos (2x)) = 4 cos (2x)
(d)  15 e−5sds

Solution: To find the integral   15 e−5sds we use the formula:

eax dx = 1/a  eax

with a = −5

 ∫   15 e−5s ds = 15   ∫   e−5s = 15 × (− 1/5 e−5s) = −3 e−5s + C

and indeed:

d/ds (−3 e−5s + C) = −3 d/ds e−5s   =  −3 × (−5 e−5s) = 15 e−5s
(e) 3/w dw

Solution: To find the integral 3/w dw we use the formula:

1/x dx = ln (x)

Thus we have:

 ∫ 3/wdw = ∫  3 × 1/wdw = 3 ∫  1/wdw = 3 ln (w) + C

This can be checked as follows:

d/dw (3 ln (w) + C) = 3d/dwln (w) = 3 × 1/w = 3/w
(f)   (es + e−s)ds

Solution: To find the integral   (es + e−s)ds we use the sum rule for integrals, rewriting it as the sum of two integrals:

(es + e−s)ds =   esds +   e−s ds

and then use:

eaxdx = 1/aeax dx

Take a = 1 in the first integral and a = −1 in the second integral. This implies:

 ∫ (es + e−s)ds = ∫ esds +  ∫ e−sds = es + ( 1/−1 ) e−s + C = es − e−s + C

Click on questions to reveal their solutions

Quiz 5: Select the indefinite integral of 4 sin (5x) + 5 cos (3x):

(a)20 cos (5x) − 15 sin (3x) + C Incorrect - please try again!
(b)4 sin ( 5x2/2 ) + 5 cos ( 3x2/2 ) Incorrect - please try again!
(c)−  2/3 cos (5x) + 5/4 sin (3x) + C Incorrect - please try again!
(d)−  4/5 cos (5x) + 5/3 sin (3x) + C Correct - well done

Explanation: Use the sum rule for indefinite integrals to rewrite the given expression as the sum of two integrals:

sin (axdx = −  1/a cos (ax)   and     cos (axdx = 1/a sin (ax)

we get:

 ∫   (4 sin (5x) + 5 cos (3x)) dx = 4   ∫   sin (5x) dx + 5   ∫   cos (3x) dx = 4 ×  (− 1/5)  cos (5x) + 5 × 1/3  sin (3x) = −  4/5 cos (5x) + 5/3 sin (3x) + C

This can be checked by differentiation.

Exercise 3: It may be shown (see the module on the Product and Quotient Rules) that:

d/dx [x(ln (x) − 1)] = ln (x)

From this result and the properties reviewed in the module on Logarithms calculate the following integrals (Hint: expressions like ln (2) are constants!):

(a)   ln (xdx

Solution: The given expression implies that:

ln (xdx = x[ln (x) − 1] + C

This can be checked by differentiating x[ln (x) − 1] + C using the product rule.

(b)   ln (2xdx

Solution: From the module on Logarithms we have:

ln (ax) = ln (a) + ln (x)

we then use the integral     ln (xdx = x[ln (x) − 1] + C calculated in Exercise 3(a). This gives:

 ∫   ln (2x) dx = ∫   (ln (2) + ln (x)) dx = ln (2) × ∫   1 dx +  ∫   ln (x) dx = xln (2) + x(ln (x) − 1) + C = x(ln (2) + (ln (x) − 1) + C = x(ln (2(x) − 1) + C

- where in the last line we used ln (2) + ln (x) = ln (2x).

(c)  ln (x3dx

Solution: From the module on Logarithms we have:

ln (xn) = n ln (x)

and the integral     ln (xdx = x[ln (x) − 1] + C calculated in Exercise 3(a). These give:

 ∫ ln (x3) dx = ∫ (3 ln (x)) dx = 3  ×  ∫   ln (x) dx = 3x(ln (x) − 1) + C
(d)  ln (3x2dx

Solution: Using the rules from the module on Logarithms we can rewrite the given expression as:

ln (3x2) = ln (3) + ln (x2) + 2 ln (x)

Thus:

 ∫   ln (3x2) dx = ∫   (ln (3) + 2 ln (x)) dx = ln (3) ×  ∫   1 dx + 2 ×  ∫   ln (x) dx = ln (3)x + 2x[ln (x) − 1] + C = x[ln (3) + 2 ln (x) − 2)] + C = x[ln (3(x2) − 2) + C

- where in the final expression we used the rules of logarithms.

Click on questions to reveal their solutions

# 3. Fixing Integration Constants

Example 3: Consider a rocket whose velocity in metres per second at time t seconds after launch is v = bt2, where b = 3m s−3. If at time t = 2s the rocket is at a position x = 30m away from the launch position, we can calculate its position at time ts as follows:

Velocity is the derivative of position with respect to time: v = dx/dt, so it follows that x is the integral of v ( = bt2m s−1):

x =    3t2 dt = 3 × 1/3t3 + C = t3 + C

The information that x = 30m at t = 2s, can be substituted into the above equation to find the value of C:

 30 = 23 + C 30 = 8 + C i.e. 22 = C

Thus, at time ts, the rocket is at x = t3 + 22m from the launch site.

Quiz 6: If y =   3xdx and at x = 2, it is measured that y = 4, calculate the integration constant.

(a)C = 2Incorrect - please try again!
(b)C = 4Incorrect - please try again!
(c)C = −2Correct - well done!
(d)C = 10Incorrect - please try again!

Explanation: We have:

 y = ∫ 3x dx = 3 × ∫ x dx = 3 × 1/2 x1 + 1 + C = 3/2 x2 + C

is the general solution. Substituting x = 2 and y = 4 into the above equation, the value of C is obtained:

 4 = 3/2 (2)2 + C 4 = 6 + C C = −2

Therefore, for all x, y = 3/2 x2 − 2

Quiz 7: Find the position of an object at time t = 4s if its velocity is given by v = αt + βm s−1, where α = 2m s−2 and β = 1m s−1, and its position at t = 1s was x = 2m.

(a)12m Incorrect - please try again!
(b)24m Incorrect - please try again!
(c)0m Incorrect - please try again!
(d)20m Correct - well done!

Explanation: Since x is the integral of v:

x = v dt = (2t + 1) dt = 2 × t dt + 1 dt = t2 + t + C

The position at time t = 1s was x = 2m, so these values may be substituted into the above equation to find C:

 2 = 11 + 1 + C 2 = 2 + C i.e. 0 = C

Therefore, for all t, x = t2 + t. At t = 4s:

x = 42 + 4 = 16 + 4 = 20m.

Quiz 8: Acceleration a is the rate of change of velocity v with respect to time t, i.e. a = dv/dt.

If a ball is thrown upwards on the Earth, its acceleration is constant and approximately a = −10m s−2. If its initial velocity was 3m s−2, when does the ball stop moving upwards, i.e. at what time is its velocity zero?

(a)0.3s Correct - well done!
(b)1s Incorrect - please try again!
(c)0.7s Incorrect - please try again!
(d)0.5s Incorrect - please try again!

Explanation: Since a = dv/dt, velocity is the integral of acceleration, i.e. v = ∫adt. The acceleration of the ball is constant, a = −10m s−2, and b = 1m s−½, so that:

v = (−10) dt = −10 × dt = −10t + C

At t = 0, v = 3m s−1, so these values may be substituted into the above equation to find the constant C:

 3 = −10 × 0 + C 3 = C

Thus, v = −10t + 3 for this problem. Therefore, when v = 0:

 0 = −10t + 3 10t = 3 t = 3 ⁄ 10

The ball therefore stops moving after 0.3s.

# 4. Quiz on Indefinite Integrals

Choose the solutions from the options given

1.Which of the following is an anti-derivative with respect to x of ƒ (x) = 2 cos (3x)?
(a)2x cos (3x)
(b)−6 sin (3x)
(c) 2/3 sin (3x)
(d) 2/3 sin ( 3/2 x2)
2.What is the integral with respect to x of ƒ (x) = 11 exp (10x)?
(a) 11/10 exp (10x) + C
(b) 11 exp (5x2) + C
(c) exp (11x) + C
(d) 110 exp (10x) + C
3.If the speed of an object is given by v = bt−½m s−1 for b = 1m s−½, what is its position x at time t = 9s if the object was at x = 3m at t = 1s?
(a) 25μm2
(b) 2.5 × 10−12m2
(c) 2.5 × 10−11m2
(d) 2.5 × 10−13m2