Question

Calculus Easy Problems!!Those are easy problems but I don't know why to get start with the the numbers!!
1.If f(2)=14, f′ is continuous, and ∫(upper limit:7,lower limit:2)f′(t)dt=25, what is the value of f(7)?
2.If g(1)=−3, g(5)=9, and ∫(upper limit:5,lower limit:1)g(x)dx=−5, evaluate the integral ∫(upper limit:5,lower limit:1)xg′(x)dx.

Answer 1

it IS easy, you just don't know how to do it.
Everything is easy WHEN you know how. Stop stressing.
the answer to the definite integral is f(upper limit) - f(lower limit) = number
so f(7) - f(2) = 25, and you're given f(2) = 14. Find f(7).
there now... that's easy isn't it?

Answer 2

I got the first one!!thank you!!

Answer 3

you're welcome :)

Answer 4

For second one use integration by parts

u =x dv = g'(x)
du =dx v = g(x)

=x*g(x) -integral g(x)
integral g(x) = -5

= x*g(x) +5
evaluate x*g(x) at 1 and 5

Answer 5

loads of thanks!!