Suppose that integral from-3 to 4.5 of f(x) dx=3, integral from-3 to-0.5 of f(x) dx=3, and integral from 2 to 4.5 of f(x) dx=3

Question

Evaluate the following definite integrals:

(a) integral from-0.5 to 2 of f(x) dx= -3

(b) integral from 2 to-0.5 of 3 f(x)-3 dx= ?

Answer 1

Since integrals represent the area between a curve and the x-axis between two values, if two integrals of the same curve share a bound they can be "stacked" next to each other to make a larger integral, or vice versa. The property is as follows:

A~B f(x)dx + B~C f(x)dx = A~C f(x)dx ; read as "The integral from A to B of f(x)dx plus the integral from B to C of f(x)dx equals the integral from A to C of f(x)dx

Thus, for this problem we are looking for which combination of the given integrals can be used to create the desired bounds.

(a) We want -0.5~2 (value labeled X); we know -3~-0.5=3, we know 2~4.5=3, and we know the overall -3~4.5=3. In fact, if you put the two smaller given integrals on a number line, you will see that the piece missing in between them is exactly the region -0.5~5!

1. Therefore, -3~-0.5 + -0.5~2 + 2~4.5 = -3~4.5
2. Putting in values for the integrals: 3 + X + 3 = 3
3. Solve: X = 3 - 3 - 3 = -3

(b) This builds on the previous problem, and uses a couple extra properties:

1. A~B f(x)dx = -B~A f(x)dx ; essentially, if you integrate from "high to low" instead of "low to high", you get the negative result
2. A~B c*f(x)dx = c* A~B f(x)dx ; a constant coefficient can be factored out and applied after integration
3. A~B f(x)dx + g(x)dx = A~B f(x)dx + A~B g(x)dx ; terms added or subtracted within an integral can be split into separate integrals with the same bounds

So, we can rewrite 2~-0.5 3f(x)-3dx as follows:

1. 2~-0.5 3f(x)-3dx = -[-0.5~2 3f(x)-3dx] *invert the bounds to get what we solved for in (a)
2. = -[-0.5~2 3f(x)dx - -0.5~2 3dx] *split the integral, making sure to keep the negative applied to both
3. = -0.5~2 3f(x)dx + -0.5~2 3dx *distribute that negative
4. = 3* -0.5~2 f(x)dx + -0.5~2 3dx *factor out the 3 coefficient
5. = 3*(-3) + -0.5~2 3dx *substitute our answer from (a)
6. = -9 + [3x from -0.5 to 2] *Begin solving the definite integral. The antiderivative of 3dx is 3x,

with the given bounds

1. = -9 + [3(2) - 3(-0.5)] *apply Fundamental Theorem of Calculus to evaluate
2. = -9 + [6 + 1.5] = -9 +7.5 *finish solving
3. = -1.5

Answer 2

Suppose that integral from-3 to 4.5 of f(x) dx=3, integral from-3 to-0.5 of f(x) dx=3, and integral from 2 to 4.5 of f(x) dx=3

( -3 , 4.5) 3

( -3, -0.5) 3

( 2, 4.5) 3

Evaluate the following definite integrals:

(a) integral from -0.5 to 2 of f(x) dx= -3

(b) integral from 2 to -0.5 of 3 f(x)-3 dx= ?

Step 1

integral from -0.5 to 2 of f(x) dx = -3, therefore integral from 2 to -0.5 of f(x) dx = 3

Step 2

integral from 2 to -0.5 of f(x) dx = 3, therefore integral from 2 to -0.5 of 3 f(x) dx = 9

Step 3

integral from 2 to -0.5 of 3 dx = 3 x ( -0.5 - 2 ) = - 7.5

Step 4

Finally subtracting the integrals in Steps 2 and 3 we obtain the answer

integral from 2 to -0.5 of (3 f(x) - 3) dx = 9 - ( - 7.5) = 16.5