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The Definite Integrals & Area Under a Curve

If we let the number of rectangles used in our Riemann Sum approximation grow to infinity, we can compute the exact area under a curve.
Area=limni=1nf(xi)Δx\displaystyle \boxed{\text{Area}=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x}


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Definite Integrals

The area under the graph of a function f(x)f(x) on an interval [a,b][a,b] is given by the definite integral.
abf(x)dx=limni=1nf(xi)Δx\boxed{\int_{a}^{b}f(x)\text{d}x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i)\Delta x}

where Δx=ban\Delta x=\frac{b-a}{n} and xi=a+iΔxx_i=a+i\Delta x


Wize Tip
The indefinite integral f(x) dx\displaystyle \int f(x)\ dx is a function.
The definite integral abf(x) dx\displaystyle \int_a^b f(x) \ dx is a number.

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Example: The Definition of the Definite Integral

Describe the area calculated by the following sum and compute the area.

limn i=1n[1n(1+in)2]\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left[\frac{1}{n}\left(1+\frac{i}{n}\right)^2\right]

abf(x)dx=limni=1nf(xi)Δx\boxed{\int_{a}^{b}f(x)\text{d}x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i)\Delta x}

From the limit, we see thatΔx=1n  \Delta x=\frac{1}{n}\ \ .

Since Δx=ban\Delta x=\frac{b-a}{n}, we get that ba=1b-a=1.

Recall that xi=a+iΔxx_i=a+i\Delta x

From the limit, we also see that f(xi)=(1+in)2f\left(x_i\right)=\left(1+\frac{i}{n}\right)^2.

So, xi=1+inx_i=1+\frac{i}{n}, meaning that a=1a=1.

ba=1    b=2b-a=1 \implies b=2

Therefore, this limit describes the area under the curve f(x)=x2f\left(x\right)=x^2 on the interval [1, 2]\left[1,\ 2\right].

limn i=1n[1n(1+in)2]\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left[\frac{1}{n}\left(1+\frac{i}{n}\right)^2\right]

=limn i=1n[1n(1+2in+i2n2)]=\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left[\frac{1}{n}\left(1+\frac{2i}{n}+\frac{i^2}{n^2}\right)\right]

=limn i=1n(1n+2in2+i2n3)=\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left(\frac{1}{n}+\frac{2i}{n^2}+\frac{i^2}{n^3}\right)

=limn(nn+2n(n+1)2n2+n(n+1)(2n+1)6n3)=\displaystyle \lim_{n\rightarrow\infty}\left(\frac{n}{n}+\frac{2n(n+1)}{2n^2}+\frac{n(n+1)(2n+1)}{6n^3}\right)

=limn(1+n+1n+2n2+3n+16n2)=\displaystyle \lim_{n\rightarrow\infty}\left(1+\frac{n+1}{n}+\frac{2n^2+3n+1}{6n^2}\right)

=limn(1+1+1n+13+12n+16n2)=\displaystyle \lim_{n\rightarrow\infty}\left(1+1+\frac{1}{n}+\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}\right)

=1+1+13=73\displaystyle=1+1+\frac{1}{3}=\boxed{\frac{7}{3}}


Thus 12x2dx=73\displaystyle \int_1^2x^2dx=\frac{7}{3}
If Δx=3n\Delta x=\frac{3}{n} andxi=iΔxx_i=i\Delta x, then what is the integral that represents

limni=1nxi(1+xi)2Δx\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^nx_i\left(1+x_i\right)^2\Delta x


Practice: Definite Integrals Using Riemann Sums

Evaluate 04(2x2+3)dx\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x using Riemann sums.

Extra Practice