The most basic, and arguably the most difficult, type of evaluation is to use the formal definition of a Riemann integral.
Exact Integrals as Limits of Sums
edit
Using the definition of an integral, we can evaluate the limit as
n
{\displaystyle n}
identities . This technique is often referred to as evaluation "by definition," and can be used to find definite integrals, as long as the integrands are fairly simple. We start with definition of the integral:
∫
a
b
f
(
x
)
d
x
{\displaystyle \int \limits _{a}^{b}f(x)dx}
=
lim
n
→
∞
[
b
−
a
n
∑
k
=
1
n
f
(
x
k
∗
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]}
Then picking
x
k
∗
{\displaystyle x_{k}^{*}}
x
k
=
a
+
k
b
−
a
n
{\displaystyle x_{k}=a+k{\frac {b-a}{n}}}
=
lim
n
→
∞
[
b
−
a
n
∑
k
=
1
n
f
(
a
+
k
b
−
a
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f{\big (}a+k{\tfrac {b-a}{n}}{\big )}\right]}
In some simple cases, this expression can be reduced to a real number, which can be interpreted as the area under the curve if
f
(
x
)
{\displaystyle f(x)}
[
a
,
b
]
{\displaystyle [a,b]}
Find
∫
0
2
x
2
d
x
{\displaystyle \int \limits _{0}^{2}x^{2}dx}
∫
0
2
x
2
d
x
{\displaystyle \int \limits _{0}^{2}x^{2}dx}
=
lim
n
→
∞
[
b
−
a
n
∑
k
=
1
n
f
(
x
k
∗
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]}
=
lim
n
→
∞
[
2
n
∑
k
=
1
n
f
(
2
k
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {2k}{n}}{\big )}\right]}
=
lim
n
→
∞
[
2
n
∑
k
=
1
n
(
2
k
n
)
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}\left({\frac {2k}{n}}\right)^{2}\right]}
=
lim
n
→
∞
[
2
n
∑
k
=
1
n
4
k
2
n
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}{\frac {4k^{2}}{n^{2}}}\right]}
=
lim
n
→
∞
[
8
n
3
∑
k
=
1
n
k
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]}
=
lim
n
→
∞
[
8
n
3
⋅
n
(
n
+
1
)
(
2
n
+
1
)
6
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]}
=
lim
n
→
∞
[
4
3
⋅
2
n
2
+
3
n
+
1
n
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {4}{3}}\cdot {\frac {2n^{2}+3n+1}{n^{2}}}\right]}
=
lim
n
→
∞
[
8
3
+
4
n
+
4
3
n
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{3}}+{\frac {4}{n}}+{\frac {4}{3n^{2}}}\right]}
=
8
3
{\displaystyle ={\frac {8}{3}}}
In other cases, it is even possible to evaluate indefinite integrals using the formal definition. We can define the indefinite integral as follows:
∫
f
(
x
)
d
x
{\displaystyle \int f(x)dx}
=
∫
0
x
f
(
t
)
d
t
=
lim
n
→
∞
[
x
−
0
n
∑
k
=
1
n
f
(
t
k
∗
)
]
{\displaystyle =\int \limits _{0}^{x}f(t)dt=\lim _{n\to \infty }\left[{\frac {x-0}{n}}\sum _{k=1}^{n}f(t_{k}^{*})\right]}
=
lim
n
→
∞
[
x
n
∑
k
=
1
n
f
(
0
+
k
(
x
−
0
)
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}0+{\tfrac {k(x-0)}{n}}{\big )}\right]}
=
lim
n
→
∞
[
x
n
∑
k
=
1
n
f
(
k
x
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]}
Suppose
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
∫
0
x
f
(
t
)
d
t
{\displaystyle \int \limits _{0}^{x}f(t)dt}
=
lim
n
→
∞
[
x
n
∑
k
=
1
n
f
(
k
x
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]}
=
lim
n
→
∞
[
x
n
∑
k
=
1
n
(
k
x
n
)
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}\left({\frac {kx}{n}}\right)^{2}\right]}
=
lim
n
→
∞
[
x
n
∑
k
=
1
n
k
2
⋅
x
2
n
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}{\frac {k^{2}\cdot x^{2}}{n^{2}}}\right]}
=
lim
n
→
∞
[
x
3
n
3
∑
k
=
1
n
k
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]}
=
lim
n
→
∞
[
x
3
n
3
∑
k
=
1
n
k
2
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]}
=
lim
n
→
∞
[
x
3
n
3
⋅
n
(
n
+
1
)
(
2
n
+
1
)
6
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]}
=
lim
n
→
∞
[
x
3
n
3
⋅
2
n
3
+
3
n
2
+
n
6
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {2n^{3}+3n^{2}+n}{6}}\right]}
=
x
3
⋅
lim
n
→
∞
[
2
n
3
6
n
3
+
3
n
2
6
n
3
+
n
6
n
3
]
{\displaystyle =x^{3}\cdot \lim _{n\to \infty }\left[{\frac {2n^{3}}{6n^{3}}}+{\frac {3n^{2}}{6n^{3}}}+{\frac {n}{6n^{3}}}\right]}
=
x
3
⋅
lim
n
→
∞
[
1
3
+
1
2
n
+
1
6
n
2
]
{\displaystyle =x^{3}\cdot \lim _{n\to \infty }\left[{\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\right]}
=
x
3
⋅
(
1
3
)
{\displaystyle =x^{3}\cdot \left({\frac {1}{3}}\right)}
=
x
3
3
{\displaystyle ={\frac {x^{3}}{3}}}
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f40e7f31f18d3120cd5854df2694f3311235f55b)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f{\big (}a+k{\tfrac {b-a}{n}}{\big )}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57b06290f59d25334cae6e3aed92ee16522f02e4)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {2k}{n}}{\big )}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8553f66ff312824e700fe689c3250cb6b09d949)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}\left({\frac {2k}{n}}\right)^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ce8f80fe79da631f23a6f06327f4de672f6612d)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}{\frac {4k^{2}}{n^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f54bd303d439fc7c9bcbd25f2722da50eb20115)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eed134e3685881ee42d7a595d5f4f32286b053bf)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e50443b6c2a1c7ab40b46665298d8ed1fd1f470c)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {4}{3}}\cdot {\frac {2n^{2}+3n+1}{n^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/218c41740e24249292060210168ee62f21cdec64)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {8}{3}}+{\frac {4}{n}}+{\frac {4}{3n^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a96db66ea4877d42fbf0c92253cb7b1be9c02b9)
![{\displaystyle =\int \limits _{0}^{x}f(t)dt=\lim _{n\to \infty }\left[{\frac {x-0}{n}}\sum _{k=1}^{n}f(t_{k}^{*})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23b23ebc6e0ba82339f10e3faa642d16f27e931f)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}0+{\tfrac {k(x-0)}{n}}{\big )}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/518c03c44547cc62f59e6c7eb03c8ff4837f181c)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75b9da80feed61c6f9468e4035bb9842e475ee56)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}\left({\frac {kx}{n}}\right)^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c07b77821541c872c1ab8143e45e40a6bd106e1)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}{\frac {k^{2}\cdot x^{2}}{n^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48139ada49a79ad9fde8058eaf5532f1cd5cab02)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37d95b2b271bb3b4613e3da68589ab9ccb2e33e0)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8bb425da40536b03bad50620678a44fa8949b02)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {2n^{3}+3n^{2}+n}{6}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/736cb2ee176fb6a1b5aa882df8b6df53a270cf88)
![{\displaystyle =x^{3}\cdot \lim _{n\to \infty }\left[{\frac {2n^{3}}{6n^{3}}}+{\frac {3n^{2}}{6n^{3}}}+{\frac {n}{6n^{3}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e542a61937ccfd11fdfe054680a08e47f9ea8da)
![{\displaystyle =x^{3}\cdot \lim _{n\to \infty }\left[{\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e95f05d13bb1bdc37fa98d99ff8e5ea8e3b46997)
