Sec 6.1 and Rules of integration.
Definition: A function F (x) is said to be an antiderivative of a function f (x) if
F ′(x) = f (x) .
Fact: A function f (x) can have more than one anti-derivative. For example x2, x2 + 1
both are antiderivatives of 2x .
Q. Given the following two sets functions match the functions in the 1st set with
their antiderivative(s) in the second set.
0, 1, 2x, 3x2 , 4x3 , 5x4 , ex + 2x
1 , ex + x2 + 2, x2 − 2, x4 + 44 , x5 − 5, x3 + π, x
Soln.
0 = (1)′, 1 = (x)′ , 2x = (x2 − 2)′, 3x2 = (x3 + π)′
4x3 = (x4 + 4)′ , 5x4 = (x5 − 5)′ , ex + 2x = (ex + x2 + 2)′
Theorem : Let G(x) be the anti-derivative of a
function f (x) , Then every anti-derivative of f (x)
must be of the form G(x) + C , for some constant C .
Q. Verify that ln (x) is the anti-derivative of the
function 1/x . Find a general expression for the
family of all anti-derivatives of the function 1/x .
Soln. (ln x)′ = 1/x . So the set of all entiderivatives
of 1/x must be {ln x + C } where C is a constant. In
particular ln x + e is an antiderivative of 1/x.
Q. Graph all the antiderivatives of 1/x
Soln. they consists of all the graphs obtained by
shifting the graph of ln(x) vertically up or down by
all possible amounts.
Q. Given any function f (x) , how many
anti-derivatives of f (x) may exists.
Soln. either 0 or infinitely many. It may so happen
that a particular function may not have an
anti-derivative. But if there exists 1
antiderivative F (x) , then every function of the
form F (x) + 1, F (x) + 2, F (x) + .001, ... is also an
antiderivative of f (x) . So in this case there would
exists infinitely many distinct antiderivatives.
We use the following symbol:
∫ f (x)dx = F (x) + C
When F (x) is the antiderivative of f (x)
The indefinite integral of f (x) is the family of all
antiderivatives denoted by F (x) + C . C is called
the constant of integration. f (x) is called the
integrand.
Rule 1. ∫ k dx = kx + C
Q. Find the indefinte integrals of the constant
4
functions 1,e, π , 3 .
Soln. x + C , ex + C , πx + C , 81x + C
Q. Find the indefinite integral of the constant
function 0.
Soln. 0x + C = C . i.e the indefinte integral of the
0 function are the constant functions.
Rule 2. ∫ xn dx =
Rule 3.
xn+1
n+1
+ C , when n =/ − 1
∫ 1x dx = ln|x| + C
Rule 4. ∫ ex dx = ex + C
Q. Find
Soln.
∫ x1 dx
x−1/2
−1/2
3/2
+ C = − √2x + C
Rule 5. ∫ k f(x) dx = k F (x) + C
Q.
∫ 2x dx = 2ln|x| + C
Q
∫ 2ex dx = 2ex + C
Q
∫ ex+ln2 dx = ∫ ex (eln 2 )dx = ∫ ex (2)dx = 2ex + C
Q
∫ ex+2 dx = ∫ ex e2 dx = e2 ex + C
Q. ∫− 3x−2 dx = (− 3x−1 / − 1) + C = 3x + C
Rule 6. ∫[f(x) ± g(x)]dx = F (x) ± G(x) + C
Note: we do not write 2 C s, one each for F (x), G(x)
respectively, but collect the constants into a
single constant.
Q. Find ∫(2x3 − 1x + 4√x − .01ex+2 ) dx
=
2x4
4
− ln|x| + 4x3/2 − .01(e2 )ex + C
3/2
Q. Find
Soln.
4
3
∫ x −3x
x dx
2
4
3
2
3
2
∫ x −3x
x dx = ∫(x − 3x)dx = x /3 − 3x /2 + C
2
Let s(t), v(t), a(t) , represent the position, velocity,
and the acceleration of a car at time t . Then the
velocity is the integral of acceleration, and
position is the integral of velocity.
Q. Let the velocity of a car is given by
v(t) = .01ex m/sec . And it is known that position of the
car s(0) at time t=0 was 0. Find the position
function s(t).
Soln. ∫ v(t) dx = s(t)
∫ .01et dt = .01et + C . So s(t) = .01et + C , for some some C.
But s(0) = .01e0 + C = .01 + C = 0 ⇒ C = − .01
Hence s(t) = .01et − .01
Initial Value problem.​ The above problem is an
example of an initial value problem. In it Typically
from the physical situation we can find some
“general solution”​ up to some undetermined
constants. Then we use the ​“initial value”​ of the
function in the given particular situation to
determine the constants and (hence) determine the
“particular solution”​ completely.
In the above example we solved the differential
equation s′(t) = .01et , with initial value s(0) = 0 to
get the general solution s(t) = .01et + C , then we used
the initial value to get the particular solution
s(t) = .01et − .01 .
Q. Solve the initial value problem:
f ′(x) = 3x2 − 4x + 8, and
f(1) = 9
Soln. The general solution is the indefinite
integral of 3x2 − 4x +8.
∫(3x2 − 4x + 8)dx = x3 − 2x2 + 8x + C is the general
solution. So f (x) = x3 − 2x2 + 8x + C , for some
particular C.
f (1) = (1)3 − 2(1)2 + 8(1) + C = 9
⇒ 7 + C = 9 ⇒ C = 2.
So f (x) = x3 − 2x2 + 8x + 2 is the particular solution
of the initial value problem.
Physical meaning of integral:
The shaded area is given by F (b) − F (a) , when F (x)
is the indefinite integral of f (x).
Q. Let the following be the graph of the velocity of
a car as function of time.
Find the distance travelled by the car from t=0, to
t=4.
Soln. The distance travelled by the car between t=0,
and t=4 is given by s(4)-s(0), where the position
function s(t) satisfies s’(t)=v(t). So s(4)-s(0) is
the area under the graph of v(t) from t=0 to t=4
which is equal to (0.25)π42 = 4π . So the distance
travelled is 4π.