IB Mathematics Applications & Interpretation Topic SL…

Integration Runs Differentiation Backwards

Differentiation takes a function to its rate of change; sometimes you know the rate and want the original.
Given the rate at which a cost grows, you may want the total cost; that reverse step is integration.
Reversing a derivative is called antidifferentiation, and it is the heart of integration.

The Power Rule In Reverse

To undo the power rule you add one to the power and divide by the new power. $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$
The rule excludes $n = -1$ because that would divide by zero; the $\dfrac{1}{x}$ case is an HL topic.
The $+C$ appears because any constant differentiates to zero, so infinitely many curves share one derivative.

Note

In The Booklet

The integral of $x^n$ and the area formula are given in the booklet.
Never drop the $+C$ on an indefinite integral.

Example question

Find $\displaystyle\int (6x^2 - 4x + 1)\, dx$.

Solution

Integrate term by term: $$\dfrac{6x^3}{3} - \dfrac{4x^2}{2} + x$$
Simplify: $$2x^3 - 2x^2 + x + C$$

Using A Boundary Condition To Pin Down C

An indefinite integral leaves the constant unknown, so one extra fact fixes which curve you want.
If you know a point the curve passes through, substitute it and solve for C.

Example

A curve has $\dfrac{dy}{dx} = 3x^2 + x$ and passes through $(0,\ 8.5)$.
Integrate: $$y = x^3 + \dfrac{x^2}{2} + C$$
Substitute the point: $8.5 = 0 + 0 + C$, so $C = 8.5$.
The curve is: $$y = x^3 + \dfrac{x^2}{2} + 8.5$$

Example question

A curve has $\dfrac{dy}{dx} = 4x^3$ and passes through $(1,\ 3)$. Find $y$.

Solution

Integrate: $$y = x^4 + C$$
Substitute: $3 = 1 + C$, so $C = 2$.

$$y = x^4 + 2$$

Definite Integrals Give The Area Under A Curve

A definite integral has limits, and for $f(x) > 0$ it gives the area between the curve and the x-axis. $$A = \int_a^b y \, dx$$
In AI you evaluate definite integrals with your GDC, but you still write the integral before reading the value.

Example

To find the area under $y = 3x^2 + 4$ from $x = 2$ to $x = 6$.

Set up: $$\displaystyle\int_2^6 (3x^2 + 4)\, dx$$
Antiderivative $x^3 + 4x$ evaluated: $$(216 + 24) - (8 + 8) = 224$$

Example question

Find the area enclosed by $y = 3x^2 + 4$, the x-axis, and the lines $x = 2$ and $x = 6$.

Solution

Write $\displaystyle\int_2^6 (3x^2 + 4)\, dx$.
Evaluate $[x^3 + 4x]_2^6 = 240 - 16$.
Area $= 224$.

Exam technique

The booklet gives $\int x^n\, dx$ and $A = \int_a^b y\, dx$; the marks are for using them.
Only indefinite integrals carry a $+C$; definite integrals give a number.
For a definite integral, write the integral set-up first, then read the value from the GDC.
The area rule assumes $f(x) > 0$ across the interval.

Active recall

State the reverse power rule for $\int x^n\, dx$, including the restriction on $n$.
Why does an indefinite integral need a $+C$?
How do you find $C$ when a boundary point is given?
Which integral gives the area under $y = f(x)$ between $x = a$ and $x = b$?

Integration Runs Differentiation Backwards

Differentiation takes a function to its rate of change; sometimes you know the rate and want the original.
Given the rate at which a cost grows, you may want the total cost; that reverse step is integration.
Reversing a derivative is called antidifferentiation, and it is the heart of integration.

The Power Rule In Reverse

To undo the power rule you add one to the power and divide by the new power. $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$
The rule excludes $n = -1$ because that would divide by zero; the $\dfrac{1}{x}$ case is an HL topic.
The $+C$ appears because any constant differentiates to zero, so infinitely many curves share one derivative.

Note

In The Booklet

The integral of $x^n$ and the area formula are given in the booklet.
Never drop the $+C$ on an indefinite integral.

Example question

Find $\displaystyle\int (6x^2 - 4x + 1)\, dx$.

Solution

Integrate term by term: $$\dfrac{6x^3}{3} - \dfrac{4x^2}{2} + x$$
Simplify: $$2x^3 - 2x^2 + x + C$$

Using A Boundary Condition To Pin Down C

An indefinite integral leaves the constant unknown, so one extra fact fixes which curve you want.
If you know a point the curve passes through, substitute it and solve for C.

Example

A curve has $\dfrac{dy}{dx} = 3x^2 + x$ and passes through $(0,\ 8.5)$.
Integrate: $$y = x^3 + \dfrac{x^2}{2} + C$$
Substitute the point: $8.5 = 0 + 0 + C$, so $C = 8.5$.
The curve is: $$y = x^3 + \dfrac{x^2}{2} + 8.5$$

Example question

A curve has $\dfrac{dy}{dx} = 4x^3$ and passes through $(1,\ 3)$. Find $y$.

Solution

Integrate: $$y = x^4 + C$$
Substitute: $3 = 1 + C$, so $C = 2$.

$$y = x^4 + 2$$

Definite Integrals Give The Area Under A Curve

A definite integral has limits, and for $f(x) > 0$ it gives the area between the curve and the x-axis. $$A = \int_a^b y \, dx$$
In AI you evaluate definite integrals with your GDC, but you still write the integral before reading the value.

Example

To find the area under $y = 3x^2 + 4$ from $x = 2$ to $x = 6$.

Set up: $$\displaystyle\int_2^6 (3x^2 + 4)\, dx$$
Antiderivative $x^3 + 4x$ evaluated: $$(216 + 24) - (8 + 8) = 224$$

Example question

Find the area enclosed by $y = 3x^2 + 4$, the x-axis, and the lines $x = 2$ and $x = 6$.

Solution

Write $\displaystyle\int_2^6 (3x^2 + 4)\, dx$.
Evaluate $[x^3 + 4x]_2^6 = 240 - 16$.
Area $= 224$.

Exam technique

The booklet gives $\int x^n\, dx$ and $A = \int_a^b y\, dx$; the marks are for using them.
Only indefinite integrals carry a $+C$; definite integrals give a number.
For a definite integral, write the integral set-up first, then read the value from the GDC.
The area rule assumes $f(x) > 0$ across the interval.

Active recall

State the reverse power rule for $\int x^n\, dx$, including the restriction on $n$.
Why does an indefinite integral need a $+C$?
How do you find $C$ when a boundary point is given?
Which integral gives the area under $y = f(x)$ between $x = a$ and $x = b$?

IB Mathematics Applications & Interpretation SL 5.5—introduction to Integration Notes

IB Mathematics Applications & Interpretation SL 5.5—introduction to Integration Notes

Integration Runs Differentiation Backwards

The Power Rule In Reverse

Using A Boundary Condition To Pin Down C

Definite Integrals Give The Area Under A Curve

Integration Runs Differentiation Backwards

The Power Rule In Reverse

Using A Boundary Condition To Pin Down C

Definite Integrals Give The Area Under A Curve

IB Mathematics Applications & Interpretation SL 5.5—introduction to Integration Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation SL 5.5—introduction to Integration Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

Integration Runs Differentiation Backwards

The Power Rule In Reverse

Using A Boundary Condition To Pin Down C

Definite Integrals Give The Area Under A Curve

Integration Runs Differentiation Backwards

The Power Rule In Reverse

Using A Boundary Condition To Pin Down C

Definite Integrals Give The Area Under A Curve