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