Unit 6 Integration

U Substitution

100

int 2xcos(x^2) dx

sin(x²)+C

200

∫1/[xln(x)]dx

ln∣ln(x)∣+C

300

∫6x(x^2+5)^4dx

(x²+5)⁵+C

400

int_1^2 xsqrt(x^2+1) dx

1/3 (5sqrt5-2sqrt2)

 

500

∫ x^2/(x^3+1)^2  dx

 

-1/[3(x^3+1)]+C

Integrating using Linear Partial Fractions

100

∫ 1/(x^2−1) dx

1/2  ln|(x-1)/(x+1)|+C 

200

∫(3x+1)/(x^2+3x)dx

1/3ln|x|+8/3ln|x+3|+C

300

∫ (5x-1)/[ (x+2)(x−3)]  dx

11/5  ln∣x+2∣+ 14/5  ln∣x−3∣+C  

400

∫ (2x+7)/[x ^2 +4x−5]  dx

1/2 ln∣x+5∣+ 3/2  ln∣x−1∣+C

500

∫ [7x^2+17x+6]/[(x+1)(x+2)(x+3)] dx

2ln|x+1|-ln|x+2|+3ln|x+3|+C

Fundamental Theorem of Calculus

100

∫ _1^4  x ^2 dx

21

200

F(x)=∫ _0^x cos(t)dt

Find F′(x).

F′(x)=cos(x)

300

F(x)=∫ _1^(x^2) ln(t)dt

Find F′(x).

4xln(x)

400

G(x)=∫_x^4 sqrt(1+t^3)dt

Find G′(x).

-sqrt(1+x^3)

500

F(x)=∫_(x^3)^(x^4) 1/(1+t^2) dt

(4x^3)/(1+x^8)-(3x^2)/(1+x^6) 

Area/Reimann Sums

100

Use geometry to find the area under f(x)=2 on the interval [0, 5]. 

Area=10

200

Estimate the area under f(x)=x from x=0 to x=4 using a left Riemann sum with 2 rectangles.

Area=4

300

Use a right Riemann sum with 3 rectangles to estimate the area under f(x)=x² on [0, 3].

Area=14

400

Approximate the area under

f(x)=sqrtx 

on [1,5] using a midpoint Riemann sum with 2 rectangles.

Area=6.828

500

f(x)=1/x

is decreasing on the interval [1,5].
Using Left Riemann sum with 4 rectangles, over or under estimate the area? Calculate the area.

Overestimate; Area=2.083

Integration by Parts

100

int xe^xdx

xe^x-e^x+C

200

int xln(x)dx

x^2/2ln(x)-x^2/4+C

300

int xcos(x)dx

xsin(x)+cos(x)+C

400

int ln(x)dx

xln(x)-x+C

500

int x^2e^xdx

x^2e^x-2xe^x+2e^x+C