evaluate the definite integrals
1
∫ root(1+3x) dx
0
= [(2/3)(1/3)(1 + 3x)^(3/2)] from 0 to 1
=[(2/9)(1 + 3x)^(3/2)] from 0 to 1
= (2/9)(4^(3/2) - (2/9)(1^(3/2))
= (2/9)(8) - (2/9)
= 14/9
1
∫ (1+3x)^.5 dx
0
let z = (1+3x)
dz = 3 dx so dx = (1/3)dz
if x = 0, z = 1
if x = 1, z = 4
so
4
∫ (z)^.5 dz/3
1
is
(1/4.5) z^1.5 at z = 4 - at z = 1
To evaluate the definite integral ∫ √(1 + 3x) dx from 0 to 1, we'll follow these steps:
Step 1: Find the antiderivative.
Step 2: Apply the fundamental theorem of calculus.
Step 3: Evaluate the integral using the limits of integration.
Let's break it down:
Step 1: Find the antiderivative.
We need to find an expression whose derivative is equal to √(1 + 3x). In this case, the antiderivative can be found using the substitution method. So, let u = 1 + 3x:
∫ √(1 + 3x) dx = ∫ √u * (1/3) dx
Next, we substitute (1/3) du for dx:
= (1/3) ∫ √u du
Now, we can integrate √u as a power rule where the exponent is 1/2:
= (1/3) * (2/3) * u^(3/2) + C
= (2/9) * u^(3/2) + C
Step 2: Apply the fundamental theorem of calculus.
To evaluate the definite integral, we will apply the fundamental theorem of calculus, which states that the definite integral is equal to the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit:
∫₀¹ √(1 + 3x) dx = [(2/9) * u^(3/2)] from 0 to 1
Step 3: Evaluate the integral using the limits of integration.
Now, substitute the upper and lower limits into the formula:
= [(2/9) * (1 + 3(1))^(3/2)] - [(2/9) * (1 + 3(0))^(3/2)]
Simplify:
= [2/9 * 4^(3/2)] - [2/9 * 1^(3/2)]
= 2/9 * 8 - 2/9 * 1
= 16/9 - 2/9
= 14/9
Therefore, the value of the definite integral is 14/9.