Explore Integrals

Iterated integrals

With definite integrals, we could calculate the area of the region between the graph of the function, the $$x$$-axis, and the limits of integration — but what if we want to calculate the area bounded by two curves? That’s where the iterated integrals come in!

Let’s take a closer look at what they are and why they’re useful.

What is the iterated integral?

For a continuous function $$f$$ on a rectangle $$R=\{(x,y)|~a\leq x\leq b{,}~c\leq y\leq d\}$$, the double integral is given by:

$$\iint\limits_{R}{f(x,y)~dA}=\int_{a}^{b}\!\!\!\int_{c}^{d}{f(x,y)~dy~dx}=\int_{c}^{d}\!\!\!\int_{a}^{b}{f(x,y)~dx~dy}$$

These kinds of integrals are called iterated integrals.

Sometimes, the order of integrals is changed. When, you ask?

The order of integration is often changed when the innermost integral in an iterated integral is too complicated to evaluate — for example, if the function in the innermost integral doesn’t have an elementary antiderivative.

Okay, so then how do we change the order of integration?

Changing the order of integration means swapping the order of variables that appear in an iterated integral.

For example, if a double integral is written so that the inner variable is $$x$$ and the outer variable is $$y$$, changing the order of integration will switch these positions so that the inner variable will become $$y$$ and the outer variable will be $$x$$.

When the order of variables changes, the limits of integration also need to be rewritten accordingly.

However, when the order of integration is changed, we still need to use the integral rules to solve it. Remember these old friends?

Why are the iterated integrals so useful?

Sometimes, we want to find the area of a region bounded by the curves of functions. Other times, we want to use the iterated integral to find the volume of a solid bounded by a positive function in the given region and the average value of the function in two variables over a given region.

For example, if continuous function $$f(x,y)$$ is defined over a rectangle $$R=[a,b]\times[c,d]$$, then the average value of the function $$f(x,y)$$ can be calculated with the formula $$f_{avg}=\frac{1}{(b-a)(d-c)}\int\int_R f(x,y)dA$$.

Cool, right?

Let’s learn how to do it!

How to solve iterated integrals

Now that we’ve got our bearings, let’s see some iterated integrals in action! We’ll walk through a few examples together.

Example 1

Find the integral:

$$\int_0^{1}\int_y^{1}\sin{(x^2)}dxdy$$

Notice that the integral should be evaluated by $$x$$ first, but this isn’t an integral from the table of integrals and cannot be solved with our usual methods. So, let’s try to change the order of integration. First, let’s write the limits of both the inner and outer integrals as inequalities:

$$y\leq x\leq1$$

$$0\leq y\leq 1$$

Notice that $$y$$ is greater than or equal to $$0$$ and less than or equal to $$x$$. Preserving the relation between the variables, rewrite the inequalities as a compound inequality:

$$0\leq y\leq x \leq1$$

Rewrite the inequality for $$y$$ in terms of $$x$$ and the inequality for $$x$$ using only constants:

$$0\leq y \leq x$$

$$0\leq x \leq 1$$

Now, change the order of integration by changing the limits and differentials:

$$\int_0^{1}\int_0^{x}\sin{(x^2)}dydx$$

To evaluate the iterated integral, we need to start by evaluating the inner indefinite integral:

$$\int\sin{(x^2)}dy$$

Since the integrand doesn’t contain $$y$$ and the integral should be evaluated for $$y$$, $$\sin{(x^2)}$$ is considered a constant in this integral and the rule $$\int adx=a\times x$$ can be used to evaluate it:

$$\sin{(x^2)}\times y$$

To evaluate the definite integral, return the limits of integration:

$$\sin{(x^2)}\times y |_0^x$$

Use $$\int_{a}^{b} f(x)dx=F(x)\big|_a^b$$ to evaluate the expression:

$$\sin{(x^2)}\times x – \sin{(x^2)}\times 0$$

Multiplying by $$0$$ results in $$0$$, so:

$$\sin{(x^2)}\times x – 0$$

Removing the zero doesn’t change the value, so let’s get rid of it:

$$\sin{(x^2)}\times x$$

Return the result in the outer integral:

$$\int_0^{1}\sin{(x^2)}\times xdx$$

To evaluate the definite integral, first evaluate the indefinite integral:

$$\int\sin{(x^2)}\times xdx$$

Look at this: by substituting $$t=x^2$$ and $$dt=2x dx$$ , the integral will match one from our table of integrals (yay!), so let’s substitute $$t=x^2$$ and $$dx=\frac{dt}{2x}$$:

$$\int\sin{(t)}\times x\times\frac{dt}{2x}$$

Cancel out the common factor $$x$$:

$$\int\sin{(t)}\times 1\times\frac{dt}{2}$$

Simplify the expression:

$$\int\frac{\sin{(t)}}{2}dt$$

Now, use the property of integral $$\int a\times f(x)dx=a\times\int f(x)dx, a\in \mathbb{R}$$:

$$\frac{1}{2}\int\sin{(t)}dt$$

Notice that this is the integral from the table of integrals, so we can use $$\int\sin{(x)}dx=-\cos{x}$$ to evaluate it:

$$\frac{1}{2}\times(-\cos{(t)})$$

Substitute back $$t=x^2$$, because the substituted values don’t represent the actual solution:

$$\frac{1}{2}\times(-\cos{(x^2)})$$

Simplify the expression:

$$-\frac{\cos{(x^2)}}{2}$$

To evaluate the definite integral, return the limits of integration:

$$-\frac{\cos{(x^2)}}{2}|_0^1$$

Use $$F(x)\big|_a^b=F(b)-F(a)$$ to evaluate the expression:

$$-\frac{\cos{(1^2)}}{2}-\left(-\frac{\cos{(0^2)}}{2}\right)$$

Evaluate the powers:

$$-\frac{\cos{(1)}}{2}-\left(-\frac{\cos{(0)}}{2}\right)$$

When there is a $$-$$ in front of an expression in parentheses, we change the sign of each term of the expression and remove the parentheses:

$$-\frac{\cos{(1)}}{2}+\frac{\cos{(0)}}{2}$$

Remember that the value of $$\cos{(0)}$$ is $$1$$, so substitute that value into the expression:

$$-\frac{\cos{(1)}}{2}+\frac{1}{2}$$

$$\frac{-\cos{(1)}+1}{2}$$

Use $$\frac{1-\cos{(t)}}{2}=\sin{\left(\frac{t}2\right)}^2$$ to simplify the expression:

$$\sin{\left(\frac{1}2\right)}^2$$

We did it! The iterated integral $$\int_0^{1}\int_y^{1}\sin{(x^2)}dxdy$$ is equal to:

$$\sin{\left(\frac{1}2\right)}^2$$

Example 2

Find the integral:

$$\int_{0}^2\int_x^{2} e^{-y^2}dydx$$

Notice that the integral should be evaluated by $$y$$ first, but this isn’t an integral from the table of integrals and can’t be solved with one of our typical methods. So, let’s try changing the order of integration! First off, let’s write the limits of both the inner and outer integrals as inequalities:

$$x\leq y\leq2$$

$$0\leq x\leq 2$$

Notice that $$x$$ is greater than or equal to $$0$$ and less than or equal to $$y$$. Preserving the relation between our variables, let’s rewrite the inequalities as a compound inequality:

$$0\leq x\leq y \leq2$$

Rewrite the inequality for $$x$$ in terms of $$y$$ and the inequality for $$y$$ using only constants:

$$0\leq x\leq y$$

$$0\leq y \leq 2$$

Now, change the order of integration by changing the limits and differentials:

$$\int_{0}^2\int_0^{y} e^{-y^2}dxdy$$

To evaluate the iterated integral, we first evaluate the inner indefinite integral:

$$\int e^{-y^2}dx$$

Since the integrand doesn’t contain $$x$$ and the integral should be evaluated for $$x$$, $$e^{-y^2}$$ is considered a constant in this integral and the rule $$\int adx=a\times x$$ can be used to evaluate it:

$$e^{-y^2}\times x$$

Use the commutative property to reorder the terms:

$$xe^{-y^2}$$

To evaluate the definite integral, return the limits of integration:

$$xe^{-y^2}|_0^y$$

Use $$F(x)\big|_a^b=F(b)-F(a)$$ to evaluate the expression:

$$ye^{-y^2}-0e^{-y^2}$$

Multiplying by $$0$$ results in $$0$$:

$$ye^{-y^2}-0$$

Removing the zero doesn’t change the value, so we can take it out:

$$ye^{-y^2}$$

Return the result in the outer integral:

$$\int_{0}^2ye^{-y^2}dy$$

To evaluate the definite integral, we need to evaluate the indefinite integral first:

$$\int ye^{-y^2}dy$$

Notice that by substituting $$t=e^{-y^2}$$ and $$dt=-2ye^{-y^2} dy$$ , the integral will be from our table of integrals! So, we’ll substitute $$t=e^{-y^2}$$ and $$dy=\frac{dt}{-2yt}$$:

$$\int yt\times \frac{dt}{-2yt}$$

Cancel out the common factors $$y$$ and $$t$$:

$$\int 1\times \frac{dt}{-2}$$

Simplify the expression:

$$\int-\frac12 dt$$

Now, use $$\int adx=a\times x$$ to evaluate the integral:

$$-\frac12 t$$

Substitute back $$t=e^{-y^2}$$ (the substituted values don’t represent the actual solution):

$$-\frac12 (e^{-y^2})$$

Simplify the expression:

$$-\frac1{2e^{(y^2)}}$$

To evaluate the definite integral, return the limits of integration:

$$-\frac1{2e^{(y^2)}}|_0^2$$

Use $$F(x)\big|_a^b=F(b)-F(a)$$ to evaluate the expression:

$$-\frac1{2e^{(2^2)}}-\left(-\frac1{2e^{(0^2)}}\right)$$

Simplify the expression:

$$-\frac{1}{2e^4}+\frac{1}{2}$$

There we have it! The iterated integral $$\int_{0}^2\int_x^{2} e^{-y^2}dydx$$ is equal to:

$$-\frac{1}{2e^4}+\frac{1}{2}$$

That wasn’t so bad, was it? You can totally do this on your own, as long as you remember these steps:

Study summary

1. If needed, change the order of integration to make the integrals easier to evaluate.
2. Evaluate the inner integral.
3. Evaluate the definite integral.

Do it yourself!

Ready to try some for yourself? We’ve got practice problems for you!

Evaluate the integral:

1. $$\int_0^{2}\int_{y^2}^{4}\sqrt{x}\sin{(x)}dxdy$$
2. $$\int_0^{8}\int_{\sqrt[3]{y}}^2e^{(x^4)} dxdy$$
3. $$\int_0^{1}\int_{\arcsin{y}}^{1}xdxdy$$
4. $$\int_0^{2}\int_{y^3}^{27}\sqrt[3]{x^2}\sin{(x)} dxdy$$

Solutions:

1. $$-4\cos{(4)}+\sin{(4)}$$
2. $$\frac{e^{16}-1}4$$
3. $$-\cos{(1)}+\sin{(1)}$$
4. $$-27\cos{(27)}+\sin{(27)}$$

If you find yourself confused or struggling, take a deep breath and get your phone (really). Open your Photomath app, scan the problem, and breathe a sigh of relief as we walk you through that tricky problem step-by-step. We’re here for you!

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