Calculus 5.2 Definite Integrals

Question

We have three different definite integrals.

Let ∫ b=9 and a=6 f(x)dx = 9, ∫ b=7 and a=6 f(x)dx=8, ∫ b=9 and a=8 f(x)dx=10

I'm asked to find two things.

Find ∫ b=8 and a=7 f(x)dx=?

and ∫ b=7 and a=8 (9f(x)-8)dx=?

I'm confused on where to start and how to solve. I would like help with both please. Thank you for your time.

Answer 1

We were given that ∫ b=9, a=6 f(x)dx = 9 , so we can actually, by laws of definite integrals split it up into pieces.

∫ b=9, a=6 f(x)dx = ∫ b=7, a=6 f(x)dx + ∫ b=8, a=7 f(x)dx + ∫ b=9, a=8 f(x)dx .

And they give you the values of each of these pieces, except ∫ b=8, a=7 f(x)dx which they want you to find.

So it's 9 = 8 + ∫ b=8, a=7 f(x)dx + 10, which means ∫ b=8, a=7 f(x)dx has to equal -9.

Next, ∫ b=7 and a=8 (9f(x)-8)dx can be rewritten so that they are switched to ∫ b=8 and a=7 . You can do this by adding a negative in front of the integral. So it is now

- [∫ b=8 and a=7 [9f(x)-8]] dx

You can split this into two integrals and distribute the negative

- ∫ b=8 and a=7 (9f(x))dx + ∫ b=8 and a=7 (8) dx

Because we figured out the value of ∫ b=8 and a=7 (f(x)) = -9, , then (-)9[-9] = 81.

Its integral is b=8 and a=7 [81x] , which is equal to 81.

The second part can be solved too, so ∫ b=8 and a=7 (- 8) dx gives us

b=8 and a=7 [8x]

which is equal to 8.

So 81 + 8 = 89. Thus, ∫ b=7 and a=8 (9f(x)-8)dx = 89.

Boy I wish Wyzant had an equation writer!