evaluate the definite integr
2
∫ 5/(3+2x) dx
0
well, one way:
let z = 3 + 2x
then
dz = 2 dx so dx = (1/2)dz
if x = 0, z = 3
if x = 2, z = 7
so
7
∫ (5/z)(1/2) dz
3
2
∫ 5/(3+2x) dx
0
= [ (5/2) ln(3 + 2x) ] from 0 to 2
= (5/2)( ln 7- ln 3)
= (5/2) ln (7/3)
To evaluate the definite integral ∫(5/(3+2x)) dx from 0 to 2, we can use the fundamental theorem of calculus.
The fundamental theorem of calculus states that if F(x) is the antiderivative of f(x) on an interval [a, b], then ∫[a, b] f(x) dx = F(b) - F(a).
To evaluate this integral, we need to find the antiderivative of 5/(3+2x). Let's begin by rewriting the function:
5/(3+2x) = 5 * 1/(3+2x)
Now, we can find the antiderivative of 1/(3+2x) using a u-substitution. Let u = 3+2x, then du = 2dx.
Substituting these values back into the integral, we get:
∫(5/(3+2x)) dx = 5 * ∫(1/u) du
This simplifies to:
∫(5/(3+2x)) dx = 5 * ln|u| + C
where C is the constant of integration.
Now, we can substitute back in the original expression for u:
∫(5/(3+2x)) dx = 5 * ln|3+2x| + C
To calculate the definite integral from 0 to 2, we substitute these bounds into the antiderivative:
∫[0, 2] 5/(3+2x) dx = 5 * ln|3+2(2)| - 5 * ln|3+2(0)|
Simplifying further:
∫[0, 2] 5/(3+2x) dx = 5 * ln|7| - 5 * ln|3|
We can now evaluate this expression:
∫[0, 2] 5/(3+2x) dx ≈ 5 * ln(7) - 5 * ln(3)
So, the approximate value of the definite integral is 5 * ln(7) - 5 * ln(3).