Assume f is continuious and has the following properties:

Question

3 10 10

f(t)dt=-2 f(t)dt=14 f(t)dt=5

0 3 8

10

a. f(t)=dt =______

0

8

b. f(t)dt

3

sqrt. 15

c. x f((1/3) x²+5) dx

3

Answer 1

Let's call the three listed integrals A, B, and C

Then,

a) the integral from 0 to 10 of f(x) is A + B = -2 + 14 = 12

b) As the integral from 3 to 8 of f(x) + C = B

the integral from 3 to 8 of f(x) equals B - C = 14 - 5 = 9

c) Let 1/3 x^2 + 5 = u

Then, du/dx = 2/3 x, or dx = 3/(2x) du

When x = 3, u = 1/3*3^2+5 = 3 + 5 = 8

When x = √15, 1/3x^2 + 5 = 1/3 * 15 + 5 = 10

Then, using I[a,b] for the integral from a to b.

I[3,√15] x f(1/3x^2 + 5) dx =

I[8, 10] x f(u) 3/(2x) du =

I[8, 10] 3/2 f(u) du =

3/2 I[8, 10] f(u) du =

3/2 C =

3/2 * 5 = 15/2 or 7 1/2