1. 0.

1. The function sin(x) has a period of 2π. The area under it from 0 to π is exactly the opposite of the area “under it” from π to 2π. They cancel out, and the answer is zero. To really understand this, note that ∫(-f(x)) dx = – ∫f(x) dx. Geometrically, the “area under” a function below the x-axis is negative, exactly minus what we see as the shaded area between the function and the x-axis.

1. 0

1. The easiest way to do this is to use geometry. The usual cos(x) has a period of 2π. Here we have cos(2x), which makes the trip twice as fast. It’s period is π, which means that on the give interval, cos(2x) goes from 1 to -1 and back up to 1. The areas above and below exactly cancel out, and the answer is 0. (See solution to the first problem on this page for a more detailed explanation of “negative” areas.) The question can also be answered algebraically. The derivative of cos(2x) is 2cos(2x). We can take F(x) = cos(2x)/2, and evaluate the integral as F(π) – F(0) = cos(2π)/2 – cos(0)/2 = 1/2 – 1/2 = 0

1. 1 – 1/e

1. Using the chain rule tells us that the derivative of e^-x is -e^-x. Our anti-derivative can be F(x)= -e^-x, and we get the value F(1) – F(0) = -e^-1 – (-e^-0) = -1/e + 1 = 1 – 1/e.

1. sin(1)

1. This is a bit tricky. Notice that we have cos(x2), multiplied by 2x, which is the derivative of x2. In other words, we have here cos(g)g’ where g = x2. This suggests that we should try sin(g): (sin(x2))’ = 2x cos(x2) This is it then. We can use F(x) = sin(x2), and the answer is F(1) – F(0) = sin(1) – sin(0) = sin(1) Note: In the “Integration by substitution” module, we will discuss a formal way of doing this integral, rather than guessing the function (though that is a perfectly fine answer, if you can guess right!)

1. 1/101

1. We could multiply out (1-x)¹⁰⁰ and find the integral from polynomial integration, but that would take all day long. Let us try and think of a better way.Our function is g¹⁰⁰, where g = 1-x. Knowing how the derivatives of xⁿ behave, we should differentiate g¹⁰¹ and see what happens. The chain rule says: (g¹⁰¹)’ = 101 g¹⁰⁰ ⋅ g’ = 101 (1-x)¹⁰⁰ ⋅ (-1) = -101 (1-x)¹⁰⁰ We wanted to get out (1-x)100, so we have to multiply by -1/101, getting F(x)= -(1-x)¹⁰¹/101. The integral then evaluates to: F(1) – F(0) = -(1-1)¹⁰¹/101 – [-(1-0)¹⁰¹/101 ] = 0 + 1/101 = 1/101