Question

Can someone please help me?
I have about an 1 hour and 45 minutes left

Answer 1

By the fundamental theorem of calculus,

`\displaystyle\int_(-5)^5f'(x)\,\mathrm dx=f(5)-f(-5)`

The integral corresponds to the signed area under the graph of
`f'(x)`:

- On the interval [-5, -4], the integral is the area of a rectangle with height 5 and length 1, hence area 5*1 = 5.

- On the interval [-4, 1], the integral is the area of a right triangle with height 5 and base 5, hence area 1/2*5*5 = 25/2.

- On [1, 2], we have another triangle with height and base 1, hence area 1/2*1*1 = 1/2. But this area lies below the horizontal axis, so we take it to be negative.

- On [2, 5], we have a rectangle with height 1 and length 3, hence area 1*3 = 3, and we take it to be negative.

So, the value of the integral over [-5, 5] is

`\displaystyle\int_(-5)^5f'(x)\,\mathrm dx=5+\frac{25}2-\frac12-3=14`

We're given that
`f(5)=-1`, so

`14=-1-f(-5)\implies f(-5)=\boxed{-15}`

Answer 2

I’m gonna say -15 is the answer Fr prolly or 1