IB Math SL – Integration

IB Math SL – Integration

Questions and Answers

1. Integration, Integrals – IB Mathematics SL

How can we find the following indefinite integral?

\int (x^2-1)^4x\ dx

Solution

\int x(x^2-1)^4\ dx =\frac{1}{2} \int 2(x^2-1)^4\ dx=

=\frac{1}{2} \frac{(x^2-1)^5}{5}+c =\frac{1}{10} (x^2-1)^5+c

 

2. Integration, Integrals – IB Mathematics SL

How can we find the real number a>1 for which the following definite integral equals to 4?

\int^a_1(1+\frac{1}{x^2}) \ dx

Solution

\int^a_1(1+\frac{1}{x^2}) \ dx=4 \Rightarrow

\Rightarrow  \left[ x-\frac{1}{x} \right]_1^a=4

\Rightarrow  (a-\frac{1}{a})-(1-\frac{1}{1})=4

\Rightarrow  (a-\frac{1}{a})=4

\Rightarrow  a^2-4a-1=0

So, a=2 + \sqrt{5} since a>1

 

3. Kinematics, Integration, Integrals – IB Mathematics SL

How can we find the times at which a particle comes to rest when is moving in a straight line with acceleration 2t-6 m/sec^2 and with initial speed 8 m/sec?

Solution

The acceleration can be expressed as a=\frac{dv}{dt}

So,  \frac{dv}{dt}= 2t-6\Rightarrow

\int \ dv=\int (2t-6) \ dt \Rightarrow v=t^2-6t+c

Next, we find the constant c by setting the initial conditions of the problem (for t=0 , v=8)

v(0)=(0)^2-6(0)+c \Rightarrow 8=c.

Therefore, the velocity function is

v(t)=t^2-6t+8=(t-2)(t-4)

The particle comes to rest when v(t)=0\Rightarrow (t-2)(t-4)=0\Rightarrow

t=2\ sec \ or \ t=4 \ sec

Determine Antiderivatives

 


Basic Indefinite Integration (Polynomial, Exponential, Quotient)

 


Area Under a Quadratic Function Using Definite Integration

 


Definite Integration Involving a Basic Trig Function (above and below x-axis)


 

Application of Definite Integration (Distance)


 

Determining Area Between Two Curves – Integration Application

 


 

Area Bounded by Two Trig Functions