Problems, once again.
1. Compute the average value of:
f(x} = x/(x+3) over the interval [-a,a]
2. Find the area of the region bounded by the graph of:
y = 2√(x^2 + 1)
X axis
Y axis
Line x = 1
On the first, integrate, then divide the integral by 2a.
On the second, integrate with respect to x from x=0 to x=1
INT y dx= INT 2√(x^2 + 1) dx
1. To compute the average value of f(x) = x/(x+3) over the interval [-a,a], we need to find the definite integral of f(x) over the same interval and then divide it by 2a.
First, let's find the integral of f(x):
∫[x=-a to x=a] (x / (x + 3)) dx
To integrate this expression, you can use the method of partial fractions or use u-substitution. Let's use u-substitution.
Let u = x + 3, then du = dx.
Substituting u and du back into the integral, we have:
∫[(u-3)/u] du
Next, perform the integral of (u-3)/u:
∫[u-3]/u du = ∫(u/u) - (3/u) du
= ∫(1 - 3/u) du
= ∫1 du - ∫(3/u) du
= u - 3ln|u| + C
Now substitute back u = x + 3:
u - 3ln|u| + C = (x + 3) - 3ln|x + 3| + C
To find the definite integral over the interval [-a,a], substitute the upper and lower limits of integration:
∫[x=-a to x=a] (x / (x + 3)) dx = [(a + 3) - 3ln|a + 3| + C] - [(-a + 3) - 3ln|-a + 3| + C]
= 2a - 6ln|a + 3| + 6ln|-a + 3| + 6C
Finally, divide this result by 2a to get the average value:
Average value = [2a - 6ln|a + 3| + 6ln|-a + 3| + 6C] / (2a)
= 1 - 3ln|a + 3| + 3ln|-a + 3| + 3C/a
Therefore, the average value of f(x) = x/(x+3) over the interval [-a,a] is given by 1 - 3ln|a + 3| + 3ln|-a + 3| + 3C/a.
2. To find the area of the region bounded by the graph of y = 2√(x^2 + 1), X-axis, and Y-axis, we need to integrate the function with respect to x from the appropriate limits.
The region is bounded by the graph of y = 2√(x^2 + 1), X-axis (y = 0), Y-axis (x = 0), and the line x = 1.
To find the area, we integrate the given function with respect to x over the appropriate interval.
∫[x=0 to x=1] (2√(x^2 + 1)) dx
To evaluate this integral, we can use the power rule and the substitution method.
Let u = x^2 + 1, then du = 2x dx.
Substituting u and du back into the integral, we have:
∫[u=1 to u=2] √u du
Now, integrate √u with respect to u:
∫[u=1 to u=2] √u du = (2/3)u^(3/2) evaluated from 1 to 2
= (2/3)(2^(3/2) - 1^(3/2))
= (2/3)(2√2 - 1)
Therefore, the area of the region bounded by the graph of y = 2√(x^2 + 1), X-axis, Y-axis, and the line x = 1 is (2/3)(2√2 - 1).