Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = sin(x), the x-axis, x = 0, and x = π
a) ∫ from π to 0 sin(x)dx
b) ∫ from π to 0 -sin(x)dx
c) 2∫ from π to 0 sin(x)dx
d) 1/2∫ from π to 0 sin(x)dx
Don't trust randoms, that answer is wrong BTW
The answer is A :)
its A I took the test
The teacher gave me only these four choices. :(
clearly you want ∫ from 0 to π sin(x)dx
But that's not one of the choices.
Now recall that ∫ b to a f(x) dx = -∫ a to b f(x) dx = ∫ a to b -f(x) dx
To calculate the total area bounded by the graph of y = sin(x), the x-axis, x = 0, and x = π, we can use definite integration.
The definite integral represents the area under the curve between two given bounds. In this case, we want to find the area between the graph of y = sin(x) and the x-axis from x = 0 to x = π.
The correct integral to calculate this area is given by option c) 2∫ from π to 0 sin(x)dx. Let's break down why this is the correct choice:
Option a) ∫ from π to 0 sin(x)dx:
This integral is incorrect because the bounds are reversed. The integral should be evaluated from the smaller bound (0) to the larger bound (π).
Option b) ∫ from π to 0 -sin(x)dx:
This integral is incorrect because the negative sign in front of sin(x) would reflect the graph of sin(x) below the x-axis. We're interested in finding the area above the x-axis, so the negative sign is not suitable for our purpose.
Option d) 1/2∫ from π to 0 sin(x)dx:
This integral is incorrect because multiplying the integral by 1/2 would reduce the area by half, giving us the wrong result.
Therefore, the correct integral to calculate the total area bounded by the graph y = sin(x), the x-axis, x = 0, and x = π is 2∫ from π to 0 sin(x)dx.
Oh come on!
It's (B)
Read what I gave you!