Suppose f has absolute minimum value m and absolute maximum value M. Between wha
ID: 2855964 • Letter: S
Question
Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must integral_0^2 f(x) dx lie? Which property of integrals allows you to make your conclusion? Use the properties of integrals to verify the inequality without evaluating the integrals. integral_0^4 (x^2 - 4x + 4) dx 0 integral_0^1 square root 1 + x^2 dx integral_0^1 square root 1 + x dx 2 integral_-1^1 square root 1 + x^2 dx 2 square root 2 pi/12 integral_pi/6^pi/3 sin x dx square root 3 pi/12 Use Property 8 of integrals to estimate the value of the integral. integral_0^1 x^3 dx integral_0^3 1/x + 4 dx integral_pi/4^pi/3 x^3tan x dx integral_0^2 (x^3 - 3x + 3) dx integral_-1^1 square root 1 + x^4 dx integral_pi^2 pi (x - 2 sin x) dx Use properties of integrals, together with Exercises 21 and 28, to prove the inequality. integral_1^3 square root x^4 + 1 dx 26/3 integral_0^pi/2 x sin x dx pi^2/8Explanation / Answer
[0 to 3] 1/(x +4) dx
substitute x +4 = u
==> dx = du
==> [0 to 3] 1/(x +4) dx = [0 to 3] 1/u du
==> [0 to 3] lnu since 1/x dx = lnx
Substitute back u = x+4
==> [0 to 3] ln (x +4)
From fundamental theorem of calculus
[0 to 3] ln (x +4) = ln (3 +4) - ln (0 +4)
==> ln 7 - ln 4
==> ln (7/4)
Hence [0 to 3] 1/(x +4) dx = ln (7/4) = 0.5596