Wize AP Calculus (BC) Textbook > Integration & Accumulation of Change

The Definite Integral

Popular Courses

AP Exam Prep

0:00 / 0:00

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=lim⁡n→∞∑i=1nf(xi)Δx\displaystyle \boxed{\text{Area}=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x}Area=n→∞lim​i=1∑n​f(xi​)Δx​


PAGE BREAK
Definite Integrals
The area under the graph of a function f(x)f(x)f(x) on an interval [a,b][a,b][a,b] is given by the definite integral.
∫abf(x)dx=lim⁡n→∞∑i=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}∫ab​f(x)dx=n→∞lim​i=1∑n​f(xi​)Δx​

where Δx=b−an\Delta x=\frac{b-a}{n}Δx=nb−a​ and xi=a+iΔxx_i=a+i\Delta xxi​=a+iΔx


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

0:00 / 0:00

Example: The Definition of the Definite Integral
Describe the area calculated by the following sum and compute the area. 

 lim⁡n→∞ ∑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]n→∞lim​ i=1∑n​[n1​(1+ni​)2] 

∫abf(x)dx=lim⁡n→∞∑i=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}∫ab​f(x)dx=n→∞lim​i=1∑n​f(xi​)Δx​

From the limit, we see thatΔx=1n  \Delta x=\frac{1}{n}\ \ Δx=n1​  .

Since Δx=b−an\Delta x=\frac{b-a}{n}Δx=nb−a​, we get that b−a=1b-a=1b−a=1.

Recall that xi=a+iΔxx_i=a+i\Delta xxi​=a+iΔx 

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

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

 b−a=1  ⟹  b=2b-a=1 \implies b=2b−a=1⟹b=2

Therefore, this limit describes the area under the curve f(x)=x2f\left(x\right)=x^2f(x)=x2 on the interval [1, 2]\left[1,\ 2\right][1, 2].

lim⁡n→∞ ∑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]n→∞lim​ i=1∑n​[n1​(1+ni​)2]

=lim⁡n→∞ ∑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]=n→∞lim​ i=1∑n​[n1​(1+n2i​+n2i2​)]

=lim⁡n→∞ ∑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)=n→∞lim​ i=1∑n​(n1​+n22i​+n3i2​)

=lim⁡n→∞(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) =n→∞lim​(nn​+2n22n(n+1)​+6n3n(n+1)(2n+1)​)

=lim⁡n→∞(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) =n→∞lim​(1+nn+1​+6n22n2+3n+1​)

=lim⁡n→∞(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) =n→∞lim​(1+1+n1​+31​+2n1​+6n21​)

=1+1+13=73\displaystyle=1+1+\frac{1}{3}=\boxed{\frac{7}{3}}=1+1+31​=37​​

Thus ∫12x2dx=73\displaystyle \int_1^2x^2dx=\frac{7}{3}∫12​x2dx=37​

If Δx=3n\Delta x=\frac{3}{n}Δx=n3​ andxi=iΔxx_i=i\Delta xxi​=iΔx, then what is the integral that represents

 lim⁡n→∞∑i=1nxi(1+xi)2Δx\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^nx_i\left(1+x_i\right)^2\Delta xn→∞lim​i=1∑n​xi​(1+xi​)2Δx

∫−12x3 dx\displaystyle \int_{-1}^2x^3\ dx∫−12​x3 dx

∫03x2(1+x)2 dx\displaystyle \int_0^3x^2\left(1+x\right)^2\ dx∫03​x2(1+x)2 dx

∫03x(1+x)2 dx\displaystyle \int_0^3x\left(1+x\right)^2\ dx∫03​x(1+x)2 dx

∫14x(1+x)2 dx\displaystyle \int_1^4x\left(1+x\right)^2\ dx∫14​x(1+x)2 dx

∫14x3 dx\displaystyle \int_1^4x^3\ dx∫14​x3 dx

I don't know

Practice: Definite Integrals Using Riemann Sums
Evaluate ∫04(2x2+3)dx\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x∫04​(2x2+3)dx using Riemann sums.