find the area under the curve y= 1/(3x+1)^2 over the interval [0,1]. please show all steps

x is + over the interval so the function we want to integrate is always + over the interval

when you see something squared on the bottom think:

d/du (v/u) = [ u dv/du - v ] / u^2
so
here let
u = (3x+1)
if v were 1
d/du(v/u) = [0-3]/(3x+1)^2
so let v = -1/3
and
integral = -(1/3)/(3x+1) + c
now of course check that by taking the derivative
now evaluate that at 1
-(1/3)/4 = -1/12 + c
evaluate that at 0
-(1/3)/1 = -1/3 + c
subtract
-1/12 - (-4/12) = 3/12 = 1/4

Sure, I can find the area under the curve, but I don't have the ability to show steps. However, I can explain the general process.

To find the area under a curve, we can use integration. In this case, the function is y = 1/(3x+1)^2.

Step 1: Determine the antiderivative of the function.
In this case, the antiderivative of y = 1/(3x+1)^2 is F(x) = -1/(3(3x+1))

Step 2: Evaluate the antiderivative at the upper and lower limits of the interval.
To find the area between x = 0 and x = 1, we will calculate F(1) - F(0).

Step 3: Plug in the values into the antiderivative and subtract.
F(1) = -1/(3(3(1)+1)) = -1/12
F(0) = -1/(3(3(0)+1)) = -1/3

Therefore, the area under the curve y = 1/(3x+1)^2 over the interval [0,1] is F(1) - F(0) = -1/12 - (-1/3) = -1/12 + 1/3 = 1/4.

So, the area under the curve is 1/4.

To find the area under the curve y = 1/(3x+1)^2 over the interval [0,1], you can use the definite integral.

Step 1: Start by finding the antiderivative of the function. The antiderivative of 1/(3x+1)^2 is -1/(3x+1).

Step 2: Next, evaluate the integral from 0 to 1 using the Fundamental Theorem of Calculus:

∫[0,1] (1/(3x+1)^2) dx = [-1/(3x+1)] [from 0 to 1]

Step 3: Substitute the upper and lower limits of integration into the antiderivative:

= [-1/(3(1)+1)] - [-1/(3(0)+1)]
= [-1/4] - [-1/1]
= -1/4 + 1
= 1 - 1/4
= 3/4

Therefore, the area under the curve y = 1/(3x+1)^2 over the interval [0,1] is 3/4.

To find the area under the curve y = 1/(3x+1)^2 over the interval [0,1], you need to evaluate the definite integral of the function over that interval.

Step 1: Start by expressing the integral of y with respect to x:
∫(0 to 1) 1/(3x+1)^2 dx

Step 2: Simplify the integral by using u-substitution. Let u = 3x+1. Then, du/dx = 3, and dx = (1/3) du.
Using these substitutions, the integral becomes:
∫(0 to 1) 1/u^2 * (1/3) du

Step 3: Rewrite the integral using the new variable:
(1/3) ∫(0 to 1) u^(-2) du

Step 4: Integrate the function u^(-2) with respect to u:
(1/3) * [-u^(-1)] from 0 to 1

Step 5: Evaluate the definite integral by plugging in the limits of integration:
(1/3) * [(-1/1) - (-1/0)]

Step 6: Simplify the expression (Note: Division by zero is undefined):
(1/3) * [(-1) - (-1/0)]

Since division by zero is undefined, the integral does not exist. Therefore, the area under the curve y = 1/(3x+1)^2 over the interval [0,1] cannot be determined.