# Integration

A LevelAQAEdexcelOCR

## Integration

Integration is the inverse of differentiation. It is represented by the symbol $\int$. Indefinite integrals, which we will look at on this page, have multiple answers because of the constant of integration. Definite integrals have limits and do not have a constant of integration – they are used to find the area under a graph.

A LevelAQAEdexcelOCR

## Integration is the Inverse of Differentiation

Integration is the inverse of differentiation. This is the Fundamental Theorem of Calculus.

Example: We know that the derivative of $x^{2}$ is $2x$. This means that the integral of $2x$ is $x^{2}$, which we write as $\int2xdx=x^{2}$. Do not forget the $dx$, it will be important in later sections.

However, this definition of integration causes a problem.

Example: Differentiate

i) $x^{3}$

ii) $x^{3}+3$

iii) $x^{3}-19$

You will notice that all of these are $3x^{2}$. So what is $\int3x^{2}dx$? Is it $x^{3}$, $x^{3}+3$ or $x^{3}-19$? Under the rule established above it is all three.

This is where the constant of integration comes in.

In fact, we write $\int3x^{2}dx=x^{3}+c$, where $c$ represents any constant, because all constants differentiate to $0$. We do this for all indefinite integrals.

A LevelAQAEdexcelOCR

## Integrating a Polynomial

To integrate something of the form $x^{n}$, add $\mathbb{1}$ to the power then divide by the new power, to get $\dfrac{1}{n+1}x^{n+1}$

$\int x^{n}dx=\dfrac{1}{n+1}x^{n+1}+c$

As with differentiation, we can integrate term by term, and the integral of $n$ times a function is $n$ times the integral of a function.

This means that we can now integrate polynomials.

Example: Integrate $x^{2}+3x+4$

$\int x^{2}+3x+4dx=\int x^{2}dx+3\int xdx+4\int1dx$

Do the integration term by term.

For $x^{2}$, we add $1$ to the power to get $3$, then divide by the new power, which is $3$, to get $\dfrac{1}{3}x^{3}$

For $x=x^{1}$, we add $1$ to the power to get $2$, then divide by the new power, which is $2$, to get $\dfrac{1}{2}x^{2}$

For $1=x^{0}$, we add $1$ to the power to get $1$, then divide by the power, which is $1$, to get $x$.

Putting it all together, and not forgetting $+c$:

\begin{aligned}\int x^{2}+3x+4dx&=\dfrac{1}{3}x^{3}+\left( 3\times\dfrac{1}{2}x^{2}\right) +\left( 4\times x\right) +c\\[1.2em]&=\dfrac{1}{3}x^{3}+\dfrac{3}{2}x^{2}+4x+c\end{aligned}

Note: Our rule works for anything of the form $x^{n}$, not just positive whole numbers, so we can integrate far more than just polynomials. Also, it does not work for $n=-1$ because this would mean dividing by $0$.

A LevelAQAEdexcelOCR

## Finding the Constant of Integration

Sometimes you will be told that an integral passes through some value, and based on this asked to find the constant of integration.

Example: The curve $y=f(x)$ passes through the point $(8,200)$ and $f'(x)=9x-5$. Find $f(x)$.

$f'(x)=9x-5$

Integrate both sides,

\begin{aligned}f(x)&=\int\left( 9x-5\right) dx\\[1.2em]&=\left( 9\times\int xdx\right) -\left( 5\times\int1dx\right) \\[1.2em]&=\left( 9\times\dfrac{1}{2}x^{2}\right) -5x+c\\[1.2em]&=\dfrac{9}{2}x^{2}-5x+c\end{aligned}

$f(8)=200$

$\left( \dfrac{9}{2}\times8^{2}\right) -\left( 5\times8\right) +c=200$

$\left( \dfrac{9}{2}\times64\right) -40+c=200$

$288-40+c=200$

$248+c=200$

$c=-48$

$f(x)=\dfrac{9}{2}x^{2}-5x-48$

A LevelAQAEdexcelOCR

## Integration Example Questions

i) $\dfrac{1}{4}x^{4}+c$

ii) $\dfrac{1}{5}x^{5}+c$

iii) $x+c$

iv) $\dfrac{1}{-1}x^{-1}+c=\dfrac{-1}{x}+c$

v) $\dfrac{1}{\left( \dfrac{3}{2}\right) }x^{\frac{3}{2}}+c=\left( \dfrac{2}{3}x^{\frac{3}{2}}\right) +c$

vi) $\dfrac{1}{\left( \dfrac{-1}{2}\right) }x^{\frac{-1}{2}}+c=-2x^{\frac{-1}{2}}+c$

i) $\left( 3\times\dfrac{1}{3}x^{3}\right) +c=x^{3}+c$

ii) $\left( \dfrac{1}{5}\times\dfrac{1}{7}x^{7}\right)+c=\dfrac{1}{35}x^{7}+c$

iii) $\left( 6\times\dfrac{1}{-3}x^{-3}\right) +c=-2x^{-3}+c$

iv) $\left( 12\times\dfrac{1}{\left( \dfrac{7}{4}\right) }x^{\frac{7}{4}}\right) +c=\dfrac{48}{7}x^{\frac{7}{4}}+c$

i) $\dfrac{1}{3}x^{3}+x^{2}+x+c$

ii) $\dfrac{1}{5}x^{5}+2x^{4}+\dfrac{2}{3}x^{3}+\dfrac{9}{2}x+6x+c$

iii) $\dfrac{2}{3}x^{\frac{3}{2}}+24x^{\frac{1}{6}}+c$

$f'(x)=3x+4$

Integrate both sides.

$f(x)=\dfrac{3}{2}x^{2}+4x+c$

$f(1)=2$

$2=\left( \dfrac{3}{2}\times1^{2}\right) +\left( 4\times1\right) +c$

$2=\dfrac{3}{2}+4+c$

$2=\dfrac{11}{2}+c$

$c=-\dfrac{7}{2}$

$f(x)=\dfrac{3}{2}x^{2}+4x-\dfrac{7}{2}$

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