8.11. VIDEOS 213

(e)∫

x7ex2dx

(f)∫

sin8 (x)dx(g)

∫sin4 (x)cos7 (x)dx

(h)∫

ex sin(3x)cos(5x)dx

(i)∫ x2+7

2x5−9x4+7x3+14x2−12x−8 dx

(j)∫ 1

1+x4 dx

(k)∫ x2+3x

(x2+x+1)2(x2+3)

dx

(l)∫ x5+2x

(x2+1)2 dx

(m)∫ 1+x2

1+3x4 dx Factor the bottom intothe product of two irreduciblequadratics to get the partial frac-tions expansion. Then stop. Theremaining details are grievous.

16. Suppose x0 ∈ (a,b) and that f is a function which has n+ 1 continuous derivativeson this interval. Consider the following.

f (x) = f (x0)+∫ x

x0

f ′ (t) dt = f (x0)+(t − x) f ′ (t) |xx0+∫ x

x0

(x− t) f ′′ (t) dt

= f (x0)+ f ′ (x0)(x− x0)+∫ x

x0

(x− t) f ′′ (t) dt.

Explain the above steps and continue the process to eventually obtain Taylor’s for-mula,

f (x) = f (x0)+n

∑k=1

f (k) (x0)

k!(x− x0)

k +1n!

∫ x

x0

(x− t)n f (n+1) (t) dt

where n! ≡ n(n−1) · · ·3 ·2 ·1 if n ≥ 1 and 0! ≡ 1.

17. In the above Taylor’s formula, use the mean value theorem for integrals to obtain theexistence of some z between x0 and x such that

f (x) = f (x0)+n

∑k=1

f (k) (x0)

k!(x− x0)

k +f (n+1) (z)(n+1)!

(x− x0)n+1 .

Hint: You might consider two cases, the case when x > x0 and the case when x < x0.

8.11 Videosvolumes

8.11.16.17.VIDEOS 213(e) [xle™ dx (1) [225 ax(f) f sin (x) dx (e+)(g) fsin4 (x) cos? (x) dx (m) f ae dx Factor the bottom into(h) fe*sin (3x) cos (5x) dx the product of two irreducible2 quadratics to get the partial frac-(i) ff DoF atte tions expansion. Then stop. Theg) f ~ dx remaining details are grievous.(k) f 43x(21) 43)Suppose xo € (a,b) and that f is a function which has n+ 1 continuous derivativeson this interval. Consider the following.Fa) =f) + [ FO d=Loo)+e—9sF Ol, + [eo FO at=F (20) +f (20) (e=a0) + fe —.F"(0) aeExplain the above steps and continue the process to eventually obtain Taylor’s for-mula,n (k) x eX£2) = F000) + YO (ery EP e—ayt pe) (9 ak=1 . * 1x0where n! =n(n—1)---3-2-lifn>1and0!=1.In the above Taylor’s formula, use the mean value theorem for integrals to obtain theexistence of some z between xo and x such thatn ¢(h) (xfs) =F(0)+ Yk=1fet) (z)(n+1)!k n+l+ (x —x0)(x — x0)Hint: You might consider two cases, the case when x > xo and the case when x < xo.8.11 Videosvolumes