1. Anti-Derivatives
If ƒ = dFdx , 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) = 12 x2:
Thus, F (x) = 12 x2 is an anti-derivative of ƒ (x) = x.
However, if C is a constant:
since the derivative of a constant is zero. The general anti-derivative of x is thus 12 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:
Explanation: To find the anti-derivative of x6 first calculate the derivative of F (x) = 17 x7. Using the basic formula:
with n = 7, we have:
| dFdx | = | ddx(17 x7 ) |
| = | 17 ddx ( x7 ) | |
| = | 17 × 7x7−1 | |
| = | x6 |
This result shows that the function F (x) = 17x7 + 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) = 1n + 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:
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:
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:
Thus, the the general anti-derivative of 3x1 ⁄ 2 is 3 × 23 x3 ⁄ 2 + C = 2x3 ⁄ 2 +C
This result may be checked by differentiating F (x) = 2x3 ⁄ 2 +C
If dFdx = ƒ (x) and dGdx = g (x), then, from the sum rule of differentiation (see module on XXX):
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?
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) = 1n + 1 for n ≠ −1, forn = 2:
and so the anti-derivative of 3x2 is:
Similarly, the anti-derivative of 2x3 is:
Putting these results together we find that the general anti-derivative of 3x2 − 2x3 is:
- 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 (x) dx + 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 ∫ x4 dx , we recall that the anti-derivatve of xn for x ≠ −1 is xn + 1 ⁄ (n + 1). Here, n = 4, so we have:
Quiz 4: Select the correct result for the indefinite integral ∫ 1√x dx
Explanation: To calculate the indefinite integral
we recall the basic result, that the anti-derivative of ƒ (x) = xn is F (x) = 1n + 1 xn+1 for n ≠ −1. In this case n = −½ and so:
| ∫ x−1 ⁄ 2dx | = | 1− ½ + 1 x− ½ + 1 + C = 1½ x½ + C |
| = | 1 × 21 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) |
| dydx | a cos (ax) | −a sin (ax) | a eax | 1x |
Exercise 1: From the above table of derivatives calculate the indefinite integrals of the following functions:
Solution: From the table we have:
This implies:
Solution: From the table we can see that if y = cos (ax), then its derivative with respect to x is:
Thus, we can conclude:
Solution: From the table of derivatives:
Thus the indefinite integral of eax is:
Solution: From the table of derivatives see that the derivative of ln (x) with respect to x is:
This implies that:
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!
Solution: We want to calculate ∫ x7dx. From the table of indefinite integrals, for any n ≠ −1,
In the case of n = 7:
Checking this:
Solution: To calculate the integral ∫ 2 sin (3x) dx we use the formula:
In our case a = 3. Thus we have:
| ∫ 2 sin (3x) dx | = | 2 ∫ sin (3x) dx = 2 × ( 13 ) cos (3x) + C |
| = | − 23 cos (3x) + C |
Checking:
Solution: To calculate the integral ∫ 4 cos (2x) dx we use the formula:
with a = 2. This yields:
| ∫ 4 cos (2x) dx | = | 4 ∫ cos (2x) dx = 4 × ( 12 ) sin (2x) + C |
It may be checked that:
Solution: To find the integral ∫ 15 e−5sds we use the formula:
with a = −5
| ∫ 15 e−5s ds | = | 15 ∫ e−5s |
| = | 15 × (− 15 e−5s) | |
| = | −3 e−5s + C |
and indeed:
Solution: To find the integral ∫3w dw we use the formula:
Thus we have:
| ∫ 3wdw | = | ∫ 3 × 1wdw | = | 3 ∫ 1wdw |
| = | 3 ln (w) + C |
This can be checked as follows:
Solution: To find the integral ∫ (es + e−s)ds we use the sum rule for integrals, rewriting it as the sum of two integrals:
and then use:
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):
Explanation: Use the sum rule for indefinite integrals to rewrite the given expression as the sum of two integrals:
we get:
| ∫ (4 sin (5x) + 5 cos (3x)) dx | = | 4 ∫ sin (5x) dx + 5 ∫ cos (3x) dx |
| = | 4 × (− 15) cos (5x) + 5 × 13 sin (3x) | |
| = | − 45 cos (5x) + 53 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:
From this result and the properties reviewed in the module on Logarithms calculate the following integrals (Hint: expressions like ln (2) are constants!):
Solution: The given expression implies that:
This can be checked by differentiating x[ln (x) − 1] + C using the product rule.
Solution: From the module on Logarithms we have:
we then use the integral ∫ ln (x) dx = 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).
Solution: From the module on Logarithms we have:
and the integral ∫ ln (x) dx = 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 |
Solution: Using the rules from the module on Logarithms we can rewrite the given expression as:
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 = dxdt, so it follows that x is the integral of v ( = bt2m s−1):
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 = ∫ 3x dx and at x = 2, it is measured that y = 4, calculate the integration constant.
Explanation: We have:
| y = ∫ 3x dx | = | 3 × ∫ x dx |
| = | 3 × 12 x1 + 1 + C | |
| = | 32 x2 + C |
is the general solution. Substituting x = 2 and y = 4 into the above equation, the value of C is obtained:
| 4 | = | 32 (2)2 + C |
| 4 | = | 6 + C |
| C | = | −2 |
Therefore, for all x, y = 32 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.
Explanation: Since x is the integral of v:
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:
Quiz 8: Acceleration a is the rate of change of velocity v with respect to time t, i.e. a = dvdt.
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?
Explanation: Since a = dvdt, velocity is the integral of acceleration, i.e. v = ∫a dt. The acceleration of the ball is constant, a = −10m s−2, and b = 1m s−½, so that:
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