IB Math HL – Integration

IB Math HL – Integration

Questions and Answers

1. Integration, Integrals – IB Mathematics HL

How can we find the following indefinite integral?

\int \frac{1}{4cos^2x+sin^2x}\ dx

Solution

Divide by cos^2x

\int \frac{1}{4cos^2x+sin^2x}\ dx =

=\int \frac{\frac{1}{cos^2x}}{4+\frac{sin^2x}{cos^2x}}\ dx

=\int \frac{sec^2x}{2^2+tan^2x}\ dx =\frac{1}{2}arctan(\frac{tanx}{2})+c

2. Integration, Integrals – IB Mathematics HL

How can we find the real number p>1 for which the following definite integral equals to 2?

\int^p_1(1+\frac{1}{x^2}) \ dx

Solution

\int^p_1(1+\frac{1}{x^2}) \ dx=2 \Rightarrow

\Rightarrow  \left[ x-\frac{1}{x} \right]_1^p=2

\Rightarrow  (p-\frac{1}{p})-(1-\frac{1}{1})=2

\Rightarrow  (p-\frac{1}{p})=2

\Rightarrow  p^2-2p-1=0

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

3. Kinematics, Integration, Integrals – IB Mathematics HL

How can we find the times at which a particle comes to rest when is moving in a straight line with acceleration 4t-22 m/sec^2 and with initial speed 60 m/sec?

Solution

We know that the acceleration a=\frac{dv}{dt}

So, \frac{dv}{dt}= 4t-22\Rightarrow

\int \ dv=\int(4t-22) \ dt \Rightarrow v=2t^2-22t+c

Next, we find the constant c by setting the initial conditions of the problem (for t=0 , v=60)

v(0)=2(0)^2-22(0)+c \Rightarrow 60=c.

Therefore, the velocity function is v(t)=2t^2-22t+60=2(t-5)(t-6)

The particle comes to rest when v(t)=0\Rightarrow 2(t-5)(t-6)=0\Rightarrow

t=5 \or\ t=6

4. Integration, Integrals – IB Mathematics HL

How can we find the following indefinite integral?

\int \frac{x+4}{\sqrt{1-x^2}}\ dx

Solution

\int \frac{x+4}{\sqrt{1-x^2}}\ dx =

=-\frac{1}{2}\int \frac{-2x}{\sqrt{1-x^2}}+4\int \frac{1}{\sqrt{1-x^2}}=

=-\frac{1}{2}\frac{(1-x^2)^{\frac{1}{2}}}{\frac{1}{2}}+4arcsin(x)+c

 

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