MATH1051 Lecture 10: Lecture 10
Review
Example #1
●Find the area under using 3 left rectangles(x) x on the interval [0, 3]f= 2+x
○x Δ = n
b−a
■ 1 = 3
3−0 = 3
3=
○ a (k)Δxxk= + − 1
■Because left rectangles, use this formula
■0 (k)(1) = + − 1
■kxk= − 1
○(x) 0f1=
■(x) 2f2=
■(x) 6f3=
○ (x)ΔxA = ∑
n
k=1
fk
■x(f(x) f(x) f(x))Δ 1+ 2+ 3
■1)(0 2 6) 8( + + =
Example #2
●Find the area under using 3 midpoint rectangles(x) x on the interval [0, ]f= − 1 6
○x 2Δ = n
b−a= 3
6−0 =
○a (k)Δxxk= + − 2
1
■0 (k)(2)= + − 2
1
■2kxk= − 1
■1, x3, x 5 x1= 2= 3=
○(x) f(1) 0f1= =
■(x) f(3) 2f2= =
■(x) f(5) 4f3= =
○ Δx(f(1) f(3) f(5))A= + +
■(2)(0 2 4) 2(6) 12= + + = =
Example #3
●Find the area under on the interval as a limit of Riemann Sums(x) xf = 2+x2, ][ 5
○xΔ = n
b−a= n
3
○a k(Δx)xk= +
■ k( )2 + n
3
Document Summary
= 2 + x x on the interval [0, 3] using 3 left rectangles f (x) ) x xk = + 1 x = n. Find the area under on the interval [0, ] Because left rectangles, use this formula (k. = + 1 xk = 1 (x ) 0 (x ) f f (x ) n. = 1 (x) x b a = 3. ) x xk = + 2 a (k. Find the area under (x) f b a = n. 3 x = n k( x) a xk = + = 2 + x x on the interval. [ 5 as a limit of riemann sums. 3k + 2 2 + n (x ) k = n. 15k + 6 (x ) k = n2 n f. A = lim (x ) x n k=1 n. 1 sin x2 cosx dx cos x2 cosx dx osxdx c.
