# How do you find the integral of #sin pi x# on the interval 0 to 1?

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To find the integral of sin(πx) on the interval [0,1], you can use integration by substitution. Let u = πx, then du = πdx. Substituting u = πx into the integral, we get:

∫ sin(πx) dx = (1/π) ∫ sin(u) du

The integral of sin(u) is -cos(u). So,

∫ sin(πx) dx = (1/π) ∫ sin(u) du = -(1/π) cos(u) + C

Substituting back u = πx, we have:

∫ sin(πx) dx = -(1/π) cos(πx) + C

Now, evaluate the definite integral from 0 to 1:

∫[0 to 1] sin(πx) dx = -(1/π) [cos(π(1)) - cos(π(0))] = -(1/π) [cos(π) - cos(0)] = -(1/π) [-1 - 1] = 2/π

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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