Sign In | Starter Of The Day | Tablesmaster | Fun Maths | Maths Map | Topics | More

International Baccalaureate Mathematics

Calculus

Syllabus Content

Derivatives of tanx, secx, cosecx, cotx, ax, logax, arcsinx, arccosx, arctanx.
Indefinite integrals of the derivatives of any of the above functions.
The composites of any of these with a linear function.
Use of partial fractions to rearrange the integrand.

Here are some specific activities, investigations or visual aids we have picked out. Click anywhere in the grey area to access the resource.

Here are some exam-style questions on this statement:

See all these questions

Click on a topic below for suggested lesson Starters, resources and activities from Transum.


Furthermore

Official Guidance, clarification and syllabus links:

Indefinite integral interpreted as a family of curves.

$$ \text{Examples } \int \dfrac{1}{x^2+2x+5}dx=\dfrac{1}{2}\arctan{\frac{(x+1)}{2}+C} $$ $$ \int \sec^2{(2x+5)}dx = \dfrac{1}{2} \tan{(2x+5)}+C$$
$$ \int \dfrac{1}{x^2+3x+2}dx = \ln|\frac{x+1}{x+2}| + C $$

Link to: partial fractions (AHL1.11)

Formula Booklet:

Standard derivatives

\( \frac{d}{dx} \tan(x) = \sec^2(x) \)

\( \frac{d}{dx} \sec(x) = \sec(x) \tan(x) \)

\( \frac{d}{dx} \cosec(x) = -\cosec(x) \cot(x) \)

\( \frac{d}{dx} \cot(x) = -\cosec^2(x) \)

\( \frac{d}{dx} a^x = a^x \ln(a) \)

\( \frac{d}{dx} \log_a(x) = \frac{1}{x \ln(a)} \)

\( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2} \)

Standard Integrals

\( \int a^x dx = \dfrac{1}{\ln{a}}a^x + C\)

\( \int \dfrac{1}{a^2+x^2} dx = \dfrac{1}{a} \arctan{\left(\frac{x}{a}\right)}+C \)

\( \int \dfrac{1}{\sqrt{a^2-x^2}} dx = \arcsin{\left(\frac{x}{a}\right)}+C, \quad |x|\lt a \)

See the main Partial Fractions section of the syllabus in Number and Algebra.

Here is an example of how to use partial fractions to decompose the rational function \( \dfrac{2x-5}{(x-2)(x+1)^2} \):

$$\begin{aligned} \frac{2x-5}{(x-2)(x+1)^2} &= \frac{A}{x-2} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \\ &= \frac{A(x+1)^2 + B(x-2)(x+1) + C(x-2)}{(x-2)(x+1)^2} \end{aligned} $$

Multiplying both sides by the common denominator gives:

$$2x-5 = A(x+1)^2 + B(x-2)(x+1) + C(x-2)$$

Substituting \(x = 2\) gives:

$$-1 = 9A$$ $$A = -\dfrac{1}{9}$$

Substituting \(x = -1\) gives:

$$-7 = -3C$$ $$C = \dfrac{7}{3}$$

Finally, substituting \(x = 0\) gives:

$$-5 = A - 2B -2C$$

Substituting A and C gives:

$$-5 = \left(-\frac{1}{9}\right) - 2B + -2\left( \frac{7}{3} \right)$$ $$B = \dfrac{1}{9}$$

Therefore, we have:

$$\frac{2x-5}{(x-2)(x+1)^2} = -\frac{1}{9(x-2)} + \frac{1}{9(x+1)} + \frac{7}{3(x+1)^2}$$

How do you teach this topic? Do you have any tips or suggestions for other teachers? It is always useful to receive feedback and helps make these free resources even more useful for Maths teachers anywhere in the world. Click here to enter your comments.


Apple

©1997-2024 WWW.TRANSUM.ORG