1. 1/2

1. The function y = 1 – x is a straight line. Its value is 1 at x = 0, and 0 at x = 1. Under it is the right triangle with vertices (0,0), (0,1), (1,0). It covers half of the unit square, and its area is 1/2.

    1. 1/3

1. We need the anti-derivative of x². Our table of anti-derivatives says that we can use F(x) = x³/3. The value of the integral is: F(1) – F(0) = 1/3 – 0/3 = 1/3.

1. ln(10)

1. Using the table of common integrals says that we can use F(x) = ln(x) as anti-derivative. The integral is F(10) – F(1) = ln(10) – ln(1), = ln(10), (since ln(1) = 0).

1. 2

1. Our table of anti-derivatives says that we can take F(x) = -cos(x). The value of the integral is: F(π) – F(0) = -cos(π) – [-cos(0)], = -(-1) + 1 = 2

1. [e³ – e^-3] / 3

    1. The derivative of e3x is 3e3x. To get e3x, we can take F(x) = e3x/3. The value of our integral is F(1) – F(-1) = e³/3 – e^-3/3 = [e³ – e^-3] /3.