Find the area between each curve and the x-axis for the given interval.
1)y = x^4 + 5 from x = 0 and x = 5
2)y = 3x^2 + 5x - 1 from x = 1 to x = 4
3)y = 4x - x^3 from x = 0 to x = 2
These are pretty easy with calculus, however, I assume you have not had integral calculus. So what have you had? Simpson's rule can be used, as well as a number of other numerical integration algorithms.
http://en.wikipedia.org/wiki/Simpson's_rule
I've had algebra, algebra 2 and advanced math
To find the area between a curve and the x-axis for a given interval, we can use definite integration.
1) For the first question, we need to find the area between the curve y = x^4 + 5 and the x-axis, from x = 0 to x = 5. Here's how you can calculate it:
- Step 1: Find the indefinite integral of the function y = x^4 + 5 with respect to x. This gives us the antiderivative F(x) = (1/5)x^5 + 5x + C.
- Step 2: Evaluate the definite integral of F(x) from x = 0 to x = 5. We can use the formula for definite integration, which states that the definite integral of a function f(x) from a to b is given by F(b) - F(a).
Let's apply the formula:
Area = F(5) - F(0)
= [(1/5)(5)^5 + 5(5) + C] - [(1/5)(0)^5 + 5(0) + C]
= [(1/5)(3125) + 25] - 0
= 625 + 25
= 650 square units
Therefore, the area between the curve y = x^4 + 5 and the x-axis, from x = 0 to x = 5, is 650 square units.
2) For the second question, we need to find the area between the curve y = 3x^2 + 5x - 1 and the x-axis, from x = 1 to x = 4. Here's how you can calculate it:
- Step 1: Find the indefinite integral of the function y = 3x^2 + 5x - 1 with respect to x. This gives us the antiderivative F(x) = x^3 + (5/2)x^2 - x + C.
- Step 2: Evaluate the definite integral of F(x) from x = 1 to x = 4.
Area = F(4) - F(1)
= [(4)^3 + (5/2)(4)^2 - (4) + C] - [(1)^3 + (5/2)(1)^2 - (1) + C]
= [(64) + (5/2)(16) - (4)] - [(1) + (5/2)(1) - (1)]
= 64 + 40 - 4 - 1 - 5/2 + 1
= 99.5 square units
Therefore, the area between the curve y = 3x^2 + 5x - 1 and the x-axis, from x = 1 to x = 4, is 99.5 square units.
3) For the third question, we need to find the area between the curve y = 4x - x^3 and the x-axis, from x = 0 to x = 2. Here's how you can calculate it:
- Step 1: Find the indefinite integral of the function y = 4x - x^3 with respect to x. This gives us the antiderivative F(x) = 2x^2 - (1/4)x^4 + C.
- Step 2: Evaluate the definite integral of F(x) from x = 0 to x = 2.
Area = F(2) - F(0)
= [2(2)^2 - (1/4)(2)^4 + C] - [2(0)^2 - (1/4)(0)^4 + C]
= [2(4) - (1/4)(16)] - [0 - 0]
= 8 - 4
= 4 square units
Therefore, the area between the curve y = 4x - x^3 and the x-axis, from x = 0 to x = 2, is 4 square units.
I hope this explanation helps you to understand how to find the area between curves and the x-axis using definite integration.