Integral calculus (general)

Integrating powers

Integrating is differentiating, backwards. First differentiating, which you already know.

Now backwards: we integrate $x^3$.

Is that right? Let's check.

Why + c?

When we differentiated, the $c$ dropped out. That happens to every fixed number.

The result is called an antiderivative. Because of $+c$ there are many of them, one for each number $c$.

The power rule

A new example. $\int x^{5}dx$ means: integrate $x^5$. The $dx$ says: the variable is $x$.

This works with every exponent. We write $n$ for any exponent.

This is the power rule:

Four examples, always the same pattern.

Remember: exponent up by 1, divide by the new exponent, $+c$ at the end.

b) $\int x^{4}dx$ (Solution: $\frac{1}{5}{x}^5+c$)

c) $\int xdx$ (Solution: $x=x^1$, so $\frac{1}{2}{x}^2+c$)

A number in front: the constant multiple rule

If there is a number in front of the power, it stays. It goes on top of the fraction.

Often you can simplify the fraction afterwards:

A number on its own is that number times $x^0$, because $x^0=1$:

This is the constant multiple rule ($k$ is a fixed number):

Remember: the number in front stays and goes on top of the fraction.

b) $\int 12x^{5}dx$ (Solution: $\frac{12}{6}{x}^6+c=2x^6+c$)

c) $\int 7dx$ (Solution: $7x+c$)

Sums: the sum rule

For a sum, you integrate each term on its own.

The check: differentiating gives back the sum.

This is the sum rule:

Three more sums, term by term:

Remember: integrate term by term, and write $+c$ once at the end.

b) $\int (x^3-4x+2)dx$ (Solution: $\frac{1}{4}{x}^4-2x^2+2x+c$)

c) $\int (5x^4+3x^2-6)dx$ (Solution: $x^5+x^3-6x+c$)

Integrating fractions and roots

A fraction with $x$ in the denominator is a power with a negative exponent: $\frac{1}{x^3}=x^{-3}$. Then the power rule works as usual.

Simplified: $\frac{1}{-2}{x}^{-2}=-\frac{1}{2}{x}^{-2}$.

Check: ${\left(-\frac{1}{2}{x}^{-2}\right)}^{\prime }=-\frac{1}{2}\cdot (-2)x^{-3}=x^{-3}$ ✓

Be careful with negative exponents. One bigger means on the number line: one step to the right.

A root is a power too. Its exponent is a fraction: $\sqrt{x}=x^{\frac{1}{2}}$.

The new exponent is $\frac{1}{2}+1=\frac{3}{2}$. Dividing by $\frac{3}{2}$ means: times $\frac{2}{3}$.

For $x^{-\frac{1}{2}}$ in the table below: $-\frac{1}{2}+1=\frac{1}{2}$, one step to the right.

Remember: first write it as a power, then integrate as usual.

b) $\int \frac{8}{x^3}dx$ (Solution: $\int 8x^{-3}dx=-4x^{-2}+c$)

c) $\int \sqrt{x^5}dx$ (Solution: $\int x^{\frac{5}{2}}dx=\frac{2}{7}{x}^{\frac{7}{2}}+c$)

d) $\int \frac{3}{\sqrt{x}}dx$ (Solution: $\int 3x^{-\frac{1}{2}}dx=6x^{\frac{1}{2}}+c$)

Basic integrals: 1/x, e to the x, sin x, cos x

There is one exponent the power rule cannot handle: $-1$.

From differentiating you know: $(\ln x{)}^{\prime }=\frac{1}{x}$, $(\sin x{)}^{\prime }=\cos x$ and $(\cos x{)}^{\prime }=-\sin x$.

First $\ln$: read backwards, $\ln$ is the antiderivative of $\frac{1}{x}$.

The absolute value $|x|$ is needed because $\ln$ exists only for positive numbers. $\frac{1}{x}$ also exists to the left of 0.

For $\sin x$ and $\cos x$ you read the circle of derivatives backwards.

The basic integrals, each with a check:

The constant multiple rule and the sum rule work here too:

Remember: you read $\frac{1}{x}$, $e^x$, $\sin x$ and $\cos x$ from the table of derivatives, backwards. The minus belongs to $\sin x$.

b) $\int (-4\sin x)dx$ (Solution: $4\cos x+c$)

c) $\int \left(6\cos x+\frac{2}{x}\right)dx$ (Solution: $6\sin x+2\ln |x|+c$)

d) $\int \left(x^3+\frac{1}{x}+\sin x\right)dx$ (Solution: $\frac{1}{4}{x}^4+\ln |x|-\cos x+c$)

Inside ax + b: divide by a

When you differentiate with the chain rule, the number from inside comes to the front. So when you integrate, you divide by it.

Check: ${\left(\frac{1}{3}{e}^{3x}\right)}^{\prime }=\frac{1}{3}\cdot 3e^{3x}=e^{3x}$ ✓

If there is also $+b$ inside, nothing changes. The $b$ drops out when you differentiate.

In the last row: outside the power rule, $\frac{1}{4}(2x+1{)}^4$, then divide by $2$.

In general ($F$ is an antiderivative of $f$):

Remember: integrate the outside as usual, then divide by $a$. The $b$ does not matter.

This only works if $ax+b$ is inside. For $e^{x^2}$ the derivative of the inside is $2x$, not a fixed number. Then you cannot divide by $2x$.

b) $\int \sin (2x-3)dx$ (Solution: $-\frac{1}{2}\cos (2x-3)+c$)

c) $\int (3x-1{)}^{4}dx$ (Solution: $\frac{1}{3}\cdot \frac{1}{5}(3x-1{)}^5+c=\frac{1}{15}(3x-1{)}^5+c$)

d) $\int \frac{1}{2x+5}dx$ (Solution: $\frac{1}{2}\ln |2x+5|+c$)

Rewrite first, then integrate

There is no separate rule for products. Integrating factor by factor is wrong: for $\int x\cdot xdx$ it would give $\frac{1}{2}{x}^2\cdot \frac{1}{2}{x}^2=\frac{1}{4}{x}^4$, but its derivative is $x^3$, not $x^2$. So rewrite first.

For a fraction with a sum on top, each term gets the denominator:

Remember: multiply out products, split fractions, then term by term.

b) $\int \frac{6x^2+3}{x}dx$ (Solution: $\int \left(6x+\frac{3}{x}\right)dx=3x^2+3\ln |x|+c$)

c) $\int \frac{2x+3}{\sqrt{x}}dx$ (Solution: $\int \left(2x^{\frac{1}{2}}+3x^{-\frac{1}{2}}\right)dx=\frac{4}{3}{x}^{\frac{3}{2}}+6x^{\frac{1}{2}}+c$)

Scheme: indefinite integral

1. Rewrite: multiply out, split fractions, write roots and fractions as powers. 2. Integrate term by term: power rule, basic integrals, for $ax+b$ divide by $a$. A number in front stays. 3. Write $+c$ once at the end. 4. Check: differentiate.

The definite integral: area under the graph

How big is the area $A$ under $f(x)=3x^2+1$ from $x=0$ to $x=2$? The top edge is curved. For a rectangle there is a formula: height times width. So we put in four rectangles, each $0.5$ wide. Their heights are $f(0)=1$, $f(0.5)=1.75$, $f(1)=4$, $f(1.5)=7.75$ and $f(2)=13$.

The sum of the rectangles with the left height is called the lower sum $U_4$, with the right height the upper sum $O_4$ (4 rectangles):

So $7.25<A<13.25$. With thinner rectangles the estimate gets better. A second example, $f(x)=x^2$ from 0 to 3, with 6, then 12, then 60 rectangles:

The sums $6.875$, $7.91$, $8.78$ approach a number: 9. This number is called the definite integral, written ${\int }_0^3{x}^{2}dx$. 0 and 3 are the limits.

You do not have to calculate it with rectangles. The antiderivative $F(x)=\frac{1}{3}{x}^3$ gives it directly: insert the upper limit, insert the lower limit, subtract. That this always works is the fundamental theorem of calculus. Here you see it: the result is exactly 9.

The area is 9 area units (AU). The first example works out too: ${\int }_0^2(3x^2+1)dx=[x^3+x{]}_0^2=10$, and 10 lies between 7.25 and 13.25.

Here you leave out the $+c$. It drops out when you subtract: $(F(b)+c)-(F(a)+c)=F(b)-F(a)$.

Scheme: definite integral

In general the limits are called $a$ (lower) and $b$ (upper).

1. Find the antiderivative $F$. Notation: $[F(x){]}_a^b$, the limits stand at the square bracket. 2. Insert the upper limit: $F(b)$. 3. Insert the lower limit: $F(a)$. 4. Subtract: $F(b)-F(a)$.

Four examples:

i) ${\int }_1^{3}2xdx=[x^2{]}_1^3=9-1=8$

ii) ${\int }_0^{2}3x^{2}dx=[x^3{]}_0^2=8-0=8$

iii) ${\int }_{-1}^2{x}^{2}dx={\left[\frac{1}{3}{x}^3\right]}_{-1}^2=\frac{8}{3}-\left(-\frac{1}{3}\right)=3$

iv) ${\int }_0^3(-2x)dx=[-x^2{]}_0^3=-9-0=-9$

In iii) the bracket matters: minus minus gives plus. In iv) the graph of $-2x$ lies below the x-axis. Then the integral is negative. An area is never negative: it is the absolute value, here 9 AU.

b) ${\int }_0^1(6x^2+2)dx$ (Solution: $[2x^3+2x{]}_0^1=4-0=4$)

c) ${\int }_0^2{e}^{x}dx$ (Solution: $[e^x{]}_0^2=e^2-1\approx 6.39$)