diff --git a/education/math/MATH1220 (calc II)/Integral Review.md b/education/math/MATH1220 (calc II)/Integral Review.md
index 351e798..8962b58 100644
--- a/education/math/MATH1220 (calc II)/Integral Review.md	
+++ b/education/math/MATH1220 (calc II)/Integral Review.md	
@@ -6,11 +6,14 @@ Divide $[a, b]$ into $n$ equal parts of width $\Delta x = \dfrac{b-a}{n}$.
 Let $x_0, x_1, x_2, \cdots, x_n$ be the endpoints of this subdivision. $x_0 = a$ and $x_n = b$. 
 
 Define $$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x$$
-- $\Delta x$ refers to the width of each sub-interval
+- $\Delta x$ refers to the width of each sub-interval, or $\frac{b-a}{n}$.
 - $f(x_i)$ refers to the height of each subinterval, and can be found with the equation $x_i = \Delta xi + a$ 
+	- Or, the width of each interval times the interval index, plus the starting offset.
 
 Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivative of $f$ (i.e $F'(x) = f(x)$).
 Then $\int_a^b f(x) dx = F(b) - F(a)$.
 
 ## Examples
-$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
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+$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
+
+$$ \int_{-2}^2 2x + 2dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$
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diff --git a/education/math/MATH1220 (calc II)/Integration by Parts.md b/education/math/MATH1220 (calc II)/Integration by Parts.md
index ae79efb..bf68eea 100644
--- a/education/math/MATH1220 (calc II)/Integration by Parts.md	
+++ b/education/math/MATH1220 (calc II)/Integration by Parts.md	
@@ -7,5 +7,11 @@ $$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
 $$\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]$$
 2. Through the distributive property of integrals,
 $$ = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
-3. Therefore:
-$$f(x)g(x) = \intf'(x)g(x)dx $$
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+3. An integral cancels out an antiderivative, therefore:
+$$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
+4. Moving terms around:
+$$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$
+
+Now, let $u = f(x)$ and $v = g(x)$, then $dv = g'(x)dx$ and $du = f'(x)dx$. 
+
+# Examples
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