The question is find the area of the reagion that is bounded by the curve y=arctan x, x=0, x=1, and the x-axis.
So I've drawn the enclosed region. To find the area would I use the Disc/shell method? If so the formula that I came up with looks like this:
If area = pi(r)^2 then it should be this...
= integral from 0 to 1 pi(arctan x)^2dx
is this correct? I took pi out to the front and then left arctanx^2.. what's the antiderivative of that? Im confused
If you want area, discs and shells don't even come into play. You only use those if you want to revolve an area around an axis.
In this case, it's a straightforward integral:
a = ∫[0,1] arctan(x) dx
= x arctan(x) - 1/2 ln(x^2+1) [0,1]
= 1/4 (π - ln(4))
To find the area of the region bounded by the curve y = arctan(x), x = 0, x = 1, and the x-axis, you need to integrate the function arctan^2(x) from x = 0 to x = 1.
Your setup is correct. The area can be calculated using the integral:
Area = ∫(0 to 1) π(arctan(x))^2 dx
To find the antiderivative of (arctan(x))^2, you can use integration techniques such as integration by parts or substitution. In this case, substitution is a good choice.
Let u = arctan(x), and du/dx = 1/(1 + x^2). Rearranging, we have dx = (1 + x^2)du. Substituting back into the integral, we get:
Area = ∫(0 to 1) π(arctan(x))^2 dx
= ∫(0 to 1) π(arctan(x))^2 (1 + x^2) du
Now, the integral is in terms of u. Expanding and simplifying, we have:
Area = π ∫(0 to 1) (u^2 + u^4) (1 + tan^2(u)) du
= π ∫(0 to 1) (u^2 + u^4 + u^6 + u^2tan^2(u)) du
Integrating each term separately, we get:
Area = π [(u^3)/3 + (u^5)/5 + (u^7)/7 + (u^3)/3tan^2(u)] |(0 to 1)
= π [(1/3 + 1/5 + 1/7 + 1/3tan^2(1)) - 0]
Simplifying further, we have:
Area = π (1/3 + 1/5 + 1/7 + 1/3tan^2(1))
Now you can evaluate the expression numerically to find the exact value of the area within the given region.
To find the area of the region bounded by the curve y = arctan(x), x = 0, x = 1, and the x-axis, you can indeed use the method of integration. The formula you have written is correct, which is:
Area = ∫[0 to 1] π(arctan(x))^2 dx
To determine the antiderivative of (arctan(x))^2, you can use integration techniques. One possible approach is to use integration by parts. Let's break it down step by step:
1. Start with the integral:
∫[0 to 1] π(arctan(x))^2 dx
2. Rewrite (arctan(x))^2 as (arctan(x))(arctan(x)).
3. Apply integration by parts using the formula:
∫u dv = uv - ∫v du
4. Choose u = arctan(x) and dv = arctan(x) dx.
5. Calculate du and v:
du = (1 / (1 + x^2)) dx
v = x * arctan(x) - ∫x (1 / (1 + x^2)) dx
6. Substitute the values of du and v into the integration by parts formula:
∫(arctan(x))^2 dx = x * arctan(x) * arctan(x) - ∫x * (1 / (1 + x^2)) dx
7. Integrate the remaining part of the equation using substitution. Let u = 1 + x^2 and du = 2x dx.
∫x * (1 / (1 + x^2)) dx = (∫du / (2u)) [substituting u]
= (1 / 2) ln(1 + x^2)
8. Finally, substitute the calculated values back into the initial equation:
∫(arctan(x))^2 dx = x * arctan(x) * arctan(x) - (1 / 2) ln(1 + x^2)
9. Evaluate the integral at the upper and lower limits:
Area = [0 to 1] x * arctan(x) * arctan(x) - (1 / 2) ln(1 + x^2)
To find the exact value of the area, you can compute the integral using a numerical method or approximate it with a calculator or software.