Question

Can someone help me with this?

Answer 1

Use the fundamental theorem of calculus
\[\int\limits_{0}^{6}f'(x) = f(6) - f(0)\]
And the definite integral also represents area under a curve
Find the area by adding up the area of the square and triangle then subtract area of small triangle below x-axis.
Set this value equal to f(6) - 7 and solve for f(6)

Answer 2

Hope that helps.
anyway answer is 12 so you can check your work.

Answer 3

hmm can you show the working out? i am kind of confused how to do.

Answer 4

explain it more please.

Answer 5

Ok find the area under the given curve.
From 0 to 2 looks like a square of length 2. Area = 4
From 2 to 4 looks like a triangle of base 2 and height 2. Area = 2
From 4 to 6 is a triangle of base 2 and height 1. Its on the neg side. Area = -1
Add up these areas. From 0 to 6 is 4+2-1 = 5
Therefore
\[\int\limits_{0}^{6}f'(x) dx = f(6) - f(0) = 5\]
Given f(0) = 7
f(6) - 7 = 5
f(6) = 12

Answer 6

Thanks!!