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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