Finding Concavity of y = Integral from x to 0

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Hello
How do I do the following
Find the interval on which the curve
y = \int_x^0 \frac{1}{1 + t + t^2} dt
is concave upward.

any help would be great.

P

Just in case Latex does not show up
x
Large S 1/1(1 + t + t^2) dt
0
 
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When, in general, is some function f concave upwards?
 
I believe a function f should be concave upward when f''(c) > 0. Am I correct. Don't have my textbook on me.
 
powp said:
I believe a function f should be concave upward when f''(c) > 0. Am I correct. Don't have my textbook on me.
Correct.

Now, if the function f (t) is y = \int_x^0 \frac{1}{1 + t + t^2} dt, then what's f'(t). (This is very easy).

Finding f''(t) is a little harder. Once found set it up as an inequality and solve.
 
f' is (x)y = \int_x^0 \frac{-1-2t}{(1 + t + t^2)^2}

f'' would then be
y = \int_x^0 \frac{-1 + 2t + 2t^2}{(1 + t + t^2)^4} dt

is this correct??

How do I solve an inequality that is this complex(it is complex to me)?? I am really not sure about this.
 
powp said:
f' is (x)y = \int_x^0 \frac{-1-2t}{(1 + t + t^2)^2}
f'' would then be
y = \int_x^0 \frac{-1 + 2t + 2t^2}{(1 + t + t^2)^4} dt
is this correct??
How do I solve an inequality that is this complex(it is complex to me)?? I am really not sure about this.
The integral is also called the anti-derivative, so the derivative of y = \int_x^0 \frac{1}{1 + t + t^2} dt
is just:
f'(t)= \frac{1}{1 + t + t^2}

So your second derivative is just:
f''(t) =\frac{-1-2t}{(1 + t + t^2)^2}
 
Not quite BobG, think about where the x is and what your actually differentiating.
 
anybody with some guidence?
 
powp said:
anybody with some guidence?
Erm, you've been given loads and loads, what have you done with what you have been given?
 
  • #10
"Fundamental Theorem of Calculus"!


What is the derivative of \int_a^x f(t)dt according to the Fundamenta Theorem of Calculus? (This is what Bobg was doing.)

Knowing that, what is the derivative of \int_x^a f(t)dt? (This is what Bobg should have done!)


What is the derivative of \int_x^a f(t)dt
 
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