Definite integrals
Okay, so you’ve learned about indefinite integrals, but now it’s the definite integral’s time to shine!
The definite integral has start and end values, unlike the indefinite integral, which has none.
Let’s see some definite integrals in action, shall we?
What is a definite integral?
Let $$f(x)$$ be a function. A definite integral
is a signed area of the region that is bounded by the function, $$x$$-axis, and the vertical lines $$x=a$$ and $$x=b$$.
Wondering how we’ll find the definite integral?
Well, to find the definite integral, we actually find the indefinite integral first and then return the limits of integration.
“Wait… but what does it mean to return the limits of integration?”
Good question!
When an indefinite integral of $$f$$ is evaluated, we get the following output:
To evaluate the definite integral of the same function over [a,b], we indicate endpoints of this interval as limits of integration (the resulting constant C can be omitted):
Just like when we worked with indefinite integrals, we’ll need to use the rules and properties of integration to help us:
| Constant multiple property of integrals | $$\int{(c\times f(x))}dx=c\times \int{f(x)}dx$$ |
| Constant multiple property of integrals | $$\int{(f(x) + g(x))}dx=\int{f(x)}dx + \int{g(x)}dx$$ |
| Difference rule for integrals | $$\int{(f(x) - g(x))}dx=\int{f(x)}dx - \int{g(x)}dx$$ |
| Substitution rule | $$\int{f(\varphi(t))}\varphi^{\prime}(t)dt=\int{f(x)}dx$$ |
| Integration by parts | $$\int{u}dv=uv-\int{v}du$$ |
Why is the definite integral so useful?
As was already mentioned, the definite integral can be used to find the signed area of the region that is bounded by the function, the $$x$$-axis, and the vertical lines $$x=a$$ and $$x=b$$, where $$a$$ and $$b$$ are the limits of integration. Say that five times fast!
For example, we have a function $$f(t)=t$$ which represents the velocity of the vehicle where $$t$$ represents the minutes. We want to know how many miles the vehicle passed in the first five minutes. We can do that by calculating the integral $$\int_0^{5} tdt$$ (spoiler alert: it’s $$12.5$$ miles).
How to find the definite integral?
Ready to work through some examples together? Here we go!
Example 1
Evaluate the integral:
To evaluate the definite integral, we’ll first evaluate the indefinite integral:
$$\int (t+1)dt$$
Notice that the expression under the integral has two terms, so use the property of integral $$\int f(x) \pm g(x)dx=\int f(x)dx\pm \int g(x)dx$$:
$$\int tdt+\int dt$$
Now we have to use the integration rules. So, use $$\int x dx=\frac{x^2}{2}$$ to evaluate the integral:
$$\frac{t^2}{2}+\int dt$$
Next, use $$\int 1 dx=x$$ to evaluate the integral:
$$\frac{t^2}{2}+t$$
To evaluate the definite integral, return the limits of integration:
$$(\frac{t^2}{2}+t)|_0^5$$
Use $$\int_{a}^{b} f(x)dx=F(b)-F(a)$$ to evaluate the expression:
$$\frac{5^2}2+5-(\frac{0^2}2+0)$$
Evaluate the expression:
$$\frac{35}2$$
There we go! The definite integral $$\int_0^{5} (t+1)dt$$ is equal to:
Example 2
Evaluate the integral:
To evaluate the definite integral, first evaluate the indefinite integral:
$$\int\frac{x^{2}}{2}dx$$
Notice that there is a constant $$\frac12$$ in the integral, so use the property $$\int a\times f(x)dx=a\times\int f(x)dx$$:
$$\frac12\int x^{2}dx$$
Now, use the rule $$\int x^{n}=\frac{x^{n+1}}{n+1}$$:
$$\frac12\times\frac{x^3}{3}$$
Multiply the fractions:
$$\frac{x^3}{6}$$
To evaluate the definite integral, return the limits of integration:
$$(\frac{x^3}{6})|_0^5$$
Use $$\int_{a}^{b} f(x)dx=F(b)-F(a)$$ to evaluate the expression:
$$\frac{1^3}6-\frac{(-1)^3}6$$
Evaluate the expression:
$$\frac13$$
Great! The definite integral $$\int_{-1}^{1}\frac{x^{2}}{2}dx$$ is equal to:
That wasn’t so bad, right? Let’s review the overall process so you can use it with any problem:
Study summary
- To evaluate the definite integral first evaluate the indefinite integral by using the properties of the integral.
- Evaluate the integral.
- To evaluate the definite integral, return the limits of integration.
- Evaluate the expression using $$F(x)\big|_a^b=F(b)-F(a)$$
- Simplify the expression.
Do it yourself!
You might not think practicing math is the most glamorous way to spend your time, but the repetition really helps cement the method in your mind. So, when you’re ready, we’ve got some practice problems for you!
Evaluate the integral:
- $$\int_0^1(5x+1)dx$$
- $$\int_{\frac{1}{2}}^1(3x^2-\ln x)dx$$
- $$\int_1^2(\cos(t)+\frac{1}{t})dt$$
- $$\int_0^1(e^x-e^{4x})dx$$
Solutions:
- $$\frac72$$
- $$\frac{11}8-\ln{(2)}\times\frac12$$
- $$\sin{2}+\ln{2}-\sin{1}$$
- $$e-\frac{e^{4}+3}{4}$$
If you’re still struggling through the solving process, that’s totally okay! Believe it or not, confusion is part of learning. If you get too stuck, just scan the problem using your Photomath app, and we’ll walk you through it!
Here’s a sneak peek of what you’ll see:
