f(x)=(x+1)/(1+x^2)
a) explain why the graph lies above the x axis on the interval x=0 and x=sqrt(3)
b)write an expression using the indefinite integral that gives the area between the graph f and the x axis between the values x=0 and x=sqrt(3).
c)use integration to find the area described in b)
for x > -1, both numerator and denominator are positive.
∫[0,√3] (x+1)/(1+x^2) dx = π/3 + ln(2)
Good luck on this one if you haven't yet studied trig substitutions.
a) To explain why the graph of f(x) lies above the x-axis on the interval x = 0 to x = sqrt(3), we need to analyze the behavior of the function and its derivative.
Let's start by finding the derivative of f(x) using the quotient rule:
f(x) = (x + 1) / (1 + x^2)
f'(x) = [(1 + x^2)(1) - (x + 1)(2x)] / (1 + x^2)^2
Simplifying, we get:
f'(x) = (1 + x^2 - 2x^2 - 2x) / (1 + x^2)^2
= -x^2 - 2x + 1 / (1 + x^2)^2
Now, let's find the critical points of f(x) by setting f'(x) equal to zero and solving for x:
-x^2 - 2x + 1 / (1 + x^2)^2 = 0
Since the denominator is always positive, we can multiply both sides by (1 + x^2)^2 without changing the sign:
-x^2 - 2x + 1 = 0
Factoring the left side:
(x + 1)(x + 1) = 0
Solving for x, we find x = -1.
Now, let's analyze the behavior of f around the critical point x = -1.
As x approaches negative infinity, both the numerator (x + 1) and the denominator (1 + x^2)^2 approach negative infinity. Therefore, the function f(x) approaches 0 from below.
As x approaches positive infinity, both the numerator (x + 1) and the denominator (1 + x^2)^2 approach positive infinity. Therefore, the function f(x) approaches 0 from above.
Since f(x) approaches 0 both from below and from above as x approaches infinity, and we know f(x) is continuous, there must be a local minimum at x = -1.
To determine whether f(x) is above or below the x-axis on the interval [0, sqrt(3)], we can evaluate f(x) at the endpoints.
f(0) = (0 + 1) / (1 + 0^2) = 1 / 1 = 1
f(sqrt(3)) = (sqrt(3) + 1) / (1 + (sqrt(3))^2) = (sqrt(3) + 1) / (1 + 3) = (sqrt(3) + 1) / 4
Since both f(0) = 1 and f(sqrt(3)) = (sqrt(3) + 1) / 4 are positive, we can conclude that the graph of f lies above the x-axis on the interval x = 0 to x = sqrt(3).
b) To write an expression for the area between the graph of f and the x-axis on the interval [0, sqrt(3)], we can use the indefinite integral. The expression will be:
∫[0, sqrt(3)] |(x + 1) / (1 + x^2)| dx
This expression represents the integral of the absolute value of f(x) over the interval [0, sqrt(3)].
c) To find the area described in part b, we need to evaluate the definite integral:
A = ∫[0, sqrt(3)] |(x + 1) / (1 + x^2)| dx
Since the function |(x + 1) / (1 + x^2)| is not easy to integrate directly, we need to use a technique called substitution to simplify the integral.
Let's substitute u = 1 + x^2. Then, du = 2x dx.
When x = 0, u = 1 + (0)^2 = 1.
When x = sqrt(3), u = 1 + (sqrt(3))^2 = 4.
Now, let's rewrite the integral using u:
A = ∫[1, 4] |(x + 1) / (1 + x^2)| du / (2x)
Since du = 2x dx, we can simplify further:
A = 1/2 ∫[1, 4] |(x + 1) / u| du
Now we have simplified the integral, and we can evaluate it by dividing it into two separate integrals:
A = 1/2 ∫[1, 4] (x + 1) / u du (when (x + 1)/u >= 0)
A = 1/2 ∫[1, 4] -(x + 1) / u du (when (x + 1)/u < 0)
By evaluating these two integrals separately, you can find the area between the graph of f and the x-axis on the interval [0, sqrt(3)].