![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2021\/12\/image-42.png\")
<\/figure>\n\n\n\nWe\u2019re back with the area under the curve, but this time, we have the interval, [katex][a, b][\/katex]. We can estimate the area of the curve by dividing it with rectangles that share a uniform width, [katex]h = \\dfrac{b – a}{n}[\/katex]. Adding these sums will lead to the area of the curve \u2013 similar to how we use the Riemann sum to estimate indefinite integrals. <\/p>\n\n\n\n
This means that when asked to calculate the area under the curve of a function, [katex]y = f(x)[\/katex], bounded by the limits, [katex]x = a[\/katex] and [katex]x =b[\/katex], we are simply evaluating the definite integral of [katex]f(x)[\/katex] between the interval of [katex]x \\in [a, b][\/katex]. <\/p>\n\n\n\n
Definite Integral Formula (Limit Sum)<\/h2>\n\n\n\n
We can define definite integrals in two ways: as an estimate of all the rectangles\u2019 sums through Riemann\u2019s sum or using the fundamental theorem of calculus relates definite and indefinite integrals. We\u2019ve briefly discussed how we can extend Riemann\u2019s sum to calculate the area of the curve and the function\u2019s definite integral, so it\u2019s time for us to establish its first formula.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{a}^{b} f(x) \\phantom{x}dx &= \\lim_{n \\rightarrow \\infty} S_n\\\\&= \\lim_{n \\rightarrow \\infty}[f(x_1)h + f(x_2)h +f(x_3)h + …+ f(x_n)h]\\\\&= \\lim_{n \\rightarrow \\infty} \\sum_{i = 1}^{n} f(x_i)h\\end{aligned} [\/katex]<\/p>\n\n\n\n
Of course, this mathematical definition is most helpful when evaluating the indefinite integral of the function is challenging. Numerical approximations of definite integrals highly depend on these approximations\u2019 formulas to calculate complex definite integrals.<\/p>\n\n\n\n
Definite Integral Formula (Fundamental Theorem of Calculus)<\/h2>\n\n\n\n
Now let\u2019s focus on a formula that\u2019s easier for us to remember and work on. By lifting the limitations for our function to be continuous as well as positive, we can now establish the mathematical definition for the function\u2019s definite integral. <\/p>\n\n\n\n
When the function, [katex]f(x)[\/katex], is bounded by the lower and upper limits: [katex]x = a[\/katex] and [katex]x =b[\/katex], respectively. <\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{a}^{b} f(x) \\phantom{x}dx &= F(b) – F(a)\\\\&= F(x)|_{a}^{b},\\\\&\\text{where } F(x) = \\int f(x) \\phantom{x}dx\\\\&\\phantom{xxxxx}F^{\\prime}(x)= f(x) \\end{aligned} [\/katex]<\/p>\n\n\n\n
This formula can be established using the fundamental theorem of calculus. It simplifies our understanding of definite and gives us a better approach to evaluating definite integrals. For the rest of our discussion, we\u2019ll focus our attention on the second formula.<\/p>\n\n\n\n
How To Evaluate Definite Integral?<\/h1>\n\n\n\n
When evaluating definite integrals, we simply evaluate the indefinite integral of the function. Suppose that we\u2019re working with [katex]\\int_{a}^{b} f(x)\\phantom{x} dx[\/katex]:<\/p>\n\n\n\n
- We simply find the expression for [katex]F(x) = \\int f(x) \\phantom{x}dx[\/katex]. <\/li>
- Evaluate [katex]F(b)[\/katex] and [katex]F(a)[\/katex].<\/li>
- Subtract these two expressions to find the definite integral of [katex]f(x)[\/katex] from [katex]x = a[\/katex] to [katex]x = b[\/katex].<\/li><\/ol>\n\n\n\n
The best way to master the process is through practice. Let\u2019s begin by trying out simple examples first.<\/p>\n\n\n\n
Problem 1<\/h3>\n\n\n\n
Evaluate the definite integral, [katex]\\int_{0}^{4} x^2 \\phantom{x}dx[\/katex].<\/p>\n\n\n\n
Let\u2019s first work on [katex]\\int x^2 \\phantom{x}dx[\/katex] by applying the power rule, [katex]\\int x^n \\phantom{x}dx = \\dfrac{x^{n + 1}}{n + 1} + C[\/katex].<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int x^2 \\phantom{x}dx &= \\dfrac{x^{2 +1}}{2 + 1} +C\\\\&= \\dfrac{x^3}{3} +C\\end{aligned} [\/katex]<\/p>\n\n\n\n
Now that we have the antiderivative of our function, [katex]F(x) = \\dfrac{x^3}{3} + C[\/katex], we simply evaluate the function at [katex]x = 0[\/katex] and [katex]x =4[\/katex].<\/p>\n\n\n\n
[katex]\\boldsymbol{x = 0}[\/katex]<\/td> [katex]\\boldsymbol{x = 4}[\/katex]<\/td><\/tr> [katex]\\begin{aligned}F(0) &= \\dfrac{0^3}{3}\\\\&= 0\\end{aligned}[\/katex]<\/td> [katex]\\begin{aligned}F(4) &= \\dfrac{4^3}{3}\\\\&= \\dfrac{64}{3}\\end{aligned}[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n To find the definite integral\u2019s value, we simply subtract these two values.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{0}^{4} x^2 \\phantom{x}dx &= \\dfrac{x^3}{3}|_{0}^{4}\\\\&= \\dfrac{64}{3} – 0\\\\&= \\dfrac{64}{3} \\end{aligned} [\/katex]<\/p>\n\n\n\n
Hence, we have [katex]\\int_{0}^{4} x^2 \\phantom{x}dx = \\dfrac{64}{3}[\/katex].<\/p>\n\n\n\n
Now, curious about what happened with the constant, [katex]C[\/katex]? Before we\u2019ve been constantly reminded to account for the arbitrary constant, [katex]C[\/katex], but since we are dealing with definite integrals, these get \u201ccanceled out\u201d.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{a}^{b} f(x) \\phantom{x}dx &= F(x) + C |_{a}^{b}\\\\&= [F(b) + C] – [F(a) + C]\\\\&= F(b) + \\cancel{C }- F(a) – \\cancel{C}\\\\&= F(b) -F(a)\\end{aligned} [\/katex]<\/p>\n\n\n\n
Of course, if we write definite integrals\u2019 evaluations this way, it becomes tedious over time. This is why moving forward; we can disregard the arbitrary constant when dealing with definite integrals.<\/p>\n\n\n\n
Definite Integral Properties and Formula<\/h2>\n\n\n\n
Before we work on more complex examples, we want to show you some of the important definite integral properties to remember. You may have encountered some of these when working with antiderivatives and indefinite integrals as well.<\/p>\n\n\n\n
1) Sum of Difference Property: <\/strong>[katex] \\boldsymbol{\\int_{a}^{b} [f(x) \\pm g(x)]\\phantom{x}dx = \\int_{a}^{b} f(x) \\phantom{x}dx \\pm \\int_{a}^{b} g(x) \\phantom{x}dx }[\/katex]<\/p>\n\n\n\n
This means that we can distribute the definite integral operation to break down complex functions.<\/p>\n\n\n\n
2) Constant Multiple Property: <\/strong>[katex] \\boldsymbol{\\int_{a}^{b} [c\\cdot f(x)]\\phantom{x}dx = c\\int_{a}^{b} f(x) \\phantom{x}dx } [\/katex]<\/p>\n\n\n\n
To easily manipulate definite integrals where the function\u2019s terms has common factors, we can factor out the common factor then evaluate the resulting function instead.<\/p>\n\n\n\n
3) Reverse Interval Property: <\/strong>[katex]\\boldsymbol{\\int_{a}^{b} f(x)\\phantom{x}dx = -\\int_{b}^{a} f(x) \\phantom{x}dx} [\/katex]<\/p>\n\n\n\n
When the limits are reversed, the resulting definite integral will be the negative of our original expression\u2019s value.<\/p>\n\n\n\n
4) Zero-Length Interval Property: <\/strong>[katex]\\boldsymbol{\\int_{a}^{b} f(x)\\phantom{x}dx = -\\int_{b}^{a} f(x) \\phantom{x}dx} [\/katex]<\/p>\n\n\n\n
When the lower and upper limits are identical, the definite integral\u2019s value will always be equal to zero.<\/p>\n\n\n\n
5) Combining Interval Property: <\/strong>[katex] \\boldsymbol{\\int_{a}^{b} f(x)\\phantom{x}dx + \\int_{b}^{c} f(x)\\phantom{x}dx = \\int_{a}^{c} f(x)\\phantom{x}dx} [\/katex]<\/p>\n\n\n\n
When given two definite integrals sharing the same lower limit and upper limit and function, we can simplify the expression into one definite integral by combining the intervals. <\/p>\n\n\n\n
These properties will come in handy when working with complex functions\u2019 definite integrals. For now, let\u2019s try applying some of these properties by trying out the problems below.<\/p>\n\n\n\n
Problem 2<\/h3>\n\n\n\n
Suppose that [katex]\\int_{-2}^{4} f(x) \\phantom{x}dx = -4[\/katex] and [katex]\\int_{-2}^{4} g(x) \\phantom{x}dx = 6[\/katex], determine the values of the following:<\/p>\n\n\n\n
a. [katex]\\int_{-2}^{4} [5f(x) \u2013 6g(x)] \\phantom{x}dx[\/katex]<\/p>\n\n\n\n
We can distribute definite integral within the expression and take out the respective factors.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{-2}^{4} [5f(x) – 6g(x)] \\phantom{x}dx &= \\int_{-2}^{4} 5f(x) \\phantom{x}dx – \\int_{-2}^{4} 6g(x) \\phantom{x}dx\\\\&= 5\\int_{-2}^{4} f(x) \\phantom{x}dx – 6\\int_{-2}^{4} g(x) \\phantom{x}dx\\end{aligned} [\/katex]<\/p>\n\n\n\n
Substitute the given values for these definite integrals to simplify the given expression.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{-2}^{4} [5f(x) – 6g(x)] \\phantom{x}dx &= 5(-4) – 6(6)\\\\&= -20 -36\\\\&= -56\\end{aligned} [\/katex]<\/p>\n\n\n\n
b. [katex]\\int_{-2}^{4} f(x) – \\int_{4}^{-2} 3g(x) \\phantom{x}dx[\/katex]<\/p>\n\n\n\n
For this problem, our second expression\u2019s lower and upper limits\u2019 positions reverse. Apply the reverse interval property to rewrite our expression.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{-2}^{4} f(x) – \\int_{4}^{-2} 3g(x) \\phantom{x}dx &=\\int_{-2}^{4} f(x) \\phantom{x}dx – \\left(-\\int_{-2}^{4} 3g(x) \\right ) \\phantom{x}dx\\\\&=\\int_{-2}^{4} f(x) \\phantom{x}dx + \\int_{-2}^{4} 3g(x) \\phantom{x}dx \\end{aligned} [\/katex]<\/p>\n\n\n\n
Now, factor out 3 in the second term and substitute appropriate values to simplify the definite integrals.<\/p>\n\n\n\n
\\begin{aligned}\\int_{-2}^{4} f(x) – \\int_{4}^{-2} 3g(x) \\phantom{x}dx &=\\int_{-2}^{4} f(x) \\phantom{x}dx + 3\\int_{-2}^{4} g(x) \\phantom{x}dx\\\\&= -4 + 3(6)\\\\&= 14 \\end{aligned}<\/p>\n\n\n\n
This means that [katex]\\int_{-2}^{4} f(x) – \\int_{4}^{-2} 3g(x) \\phantom{x}dx[\/katex] is simply equal to 14.<\/p>\n\n\n\n
Now that we\u2019ve shown you the important definite integral properties, why don\u2019t we try evaluating more complex definite integrals?<\/p>\n\n\n\n
Problem 3<\/h3>\n\n\n\n
Evaluate the definite integral, [katex]\\int_{-2}^{2} (x^3 \u2013 4x + 3) \\phantom{x}dx[\/katex].<\/p>\n\n\n\n
We can either apply the definite integral properties or work on the antiderivative first. For now, let us show you how the properties we\u2019ve discussed can be applied in evaluating definite integrals. Apply the power and constant multiple formulas for integrals when necessary. <\/p>\n\n\n\n
Method 1:<\/strong><\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{-2}^{2} (x^3 – 4x + 3) \\phantom{x}dx &= \\int_{-2}^{2} x^3 \\phantom{x}dx- \\int_{-2}^{2}4x \\phantom{x}dx+ \\int_{-2}^{2}3 \\phantom{x}dx\\\\&= \\int_{-2}^{2} x^3 \\phantom{x}dx- 4\\int_{-2}^{2}x \\phantom{x}dx+ \\int_{-2}^{2}3 \\phantom{x}dx\\\\&= \\left[\\dfrac{x^4}{4} \\right ]_{-2}^{2} -4 \\left[\\dfrac{x^2}{2} \\right ]_{-2}^{2} + [3x]_{-2}^{2}\\\\&= \\left(\\dfrac{2^4}{4} -\\dfrac{(-2)^4}{4} \\right ) -4(2^2 – (-2)^2) + 3(2 -(-2))\\\\&= 12\\end{aligned} [\/katex]<\/p>\n\n\n\n
Another way to evaluate the definite integral, [katex]\\int_{-2}^{2} (x^3 \u2013 4x + 3) \\phantom{x}dx[\/katex], is by working on its antiderivative first. Apply appropriate integral formulas when needed.<\/p>\n\n\n\n
Method 2:<\/strong><\/p>\n\n\n\n
[katex]\\begin{aligned}\\int (x^3 – 4x + 3) \\phantom{x}dx &=\\int x^3\\phantom{x}dx-\\int 4x\\phantom{x}dx + \\int 3\\phantom{x}dx\\\\&= \\int x^3\\phantom{x}dx- 4\\int x\\phantom{x}dx + \\int 3\\phantom{x}dx\\\\&= \\dfrac{x^4}{4}- 4\\cdot \\dfrac{x^2}{2}+ \\int 3\\phantom{x}dx\\\\&= \\dfrac{1}{4}x^4 – 2x^2 + 3x +C\u00a0 \\end{aligned}[\/katex]<\/p>\n\n\n\n
Once we have our function\u2019s antiderivative, we\u2019ll simply evaluate the [katex]F(x)[\/katex] at their lower and upper limits then subtract the results.<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{-2}^{2} (x^3 – 4x + 3) \\phantom{x}dx &=\\left[\\dfrac{1}{4}x^4 – 2x^2 + 3x \\right ]_{-2}^{2}\\\\&= \\left[\\dfrac{1}{4}(2)^4 – 2(2)^2 + 3(2) \\right ]-\\left[\\dfrac{1}{4}(-2)^4 – 2(-2)^2 + 3(-2) \\right ]\\\\&= 12\\end{aligned} [\/katex]<\/p>\n\n\n\n
This example shows that regardless of the process or method applied, the definite integral\u2019s value will remain the same. For the next example though, we\u2019ll only use one method. What you can do is try using the remaining method and compare your final answer!<\/p>\n\n\n\n
Problem 4<\/h3>\n\n\n\n
Evaluate the definite integral, [katex]\\int_{1}^{4} \\dfrac{x}{(x^2 + 4)^3} \\phantom{x}dx[\/katex].<\/p>\n\n\n\n
By inspecting the integrand alone, we can see that it\u2019ll be much simpler if we work on the antiderivative first. Integrating our expression will require a substitution method, but don\u2019t worry, we\u2019ll break down the steps for you here.<\/p>\n\n\n\n
[katex] \\begin{aligned}u &= x^2 + 4\\\\du &= 2x \\phantom{x}dx\\\\\\dfrac{1}{2}\\phantom{x}du &= x \\phantom{x}dx\\end{aligned} [\/katex]<\/p>\n\n\n\n
We can rewrite the integrand in terms of [katex]u[\/katex] now. Keep in mind that the limits will also change form [katex]x = 1\\Rightarrow u =5[\/katex] and [katex]x = 4\\Rightarrow u =20[\/katex].<\/p>\n\n\n\n
[katex] \\begin{aligned}\\int_{1}^{4} \\dfrac{x}{(x^2 + 4)^3} \\phantom{x}dx&= \\int_{1}^{4} \\dfrac{1}{(x^2 + 4)^3} \\phantom{x} xdx\\\\&=\\int_{5}^{20} \\dfrac{1}{2} \\dfrac{1}{u^3} \\phantom{x}du\\\\&=\\dfrac{1}{2} \\int_{5}^{8} \\dfrac{1}{u^3} \\phantom{x}du\\\\&=\\dfrac{1}{2} \\int_{5}^{20} u^{-3} \\phantom{x}du\\end{aligned} [\/katex]<\/p>\n\n\n\n
Now, simplify the antiderivative first then evaluate the expression at the new limits: [katex]u = 5[\/katex] and [katex]u = 8[\/katex].<\/p>\n\n\n\n
[katex] \\begin{aligned}\\dfrac{1}{2} \\int u^{-3} \\phantom{x}du &= \\dfrac{1}{2} \\cdot \\dfrac{u^{-3 + 1}}{-3 + 1} \\\\&= -\\dfrac{u^{-2}}{4}\\\\\\\\\\dfrac{1}{2} \\int_{5}^{20} u^{-3}&= \\left[-\\dfrac{u^{-2}}{4} \\right ]_{5}^{20}\\\\&= -\\dfrac{1}{4}[(20)^{-2} – (5)^{-2}]\\\\&= \\dfrac{3}{320}\\end{aligned} [\/katex]<\/p>\n\n\n\n
Hence, we have [katex]\\int_{1}^{4} \\dfrac{x}{(x^2 + 4)^3} \\phantom{x}dx = \\dfrac{3}{320}[\/katex].<\/p>\n","protected":false},"excerpt":{"rendered":"
In this article, we\u2019ll show you the core properties and formulas needed to evaluate and apply definite integrals.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-324","post","type-post","status-publish","format-standard","hentry","category-calculus"],"_links":{"self":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/324","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/comments?post=324"}],"version-history":[{"count":11,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/324\/revisions"}],"predecessor-version":[{"id":336,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/324\/revisions\/336"}],"wp:attachment":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/media?parent=324"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/categories?post=324"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/tags?post=324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}