<p>If f is continuous for all x, which of the following integrals necessarily have the same value?</p>
<p>I. fnInt(f(x), x, a, b)</p>
<p>II. fnInt(f(x+a),x,0,b-a)</p>
<p>III. fnInt(f(x+c),x,a+c,b+c)</p>
<p>Thanks!</p>
<p>I & III.
The “c” basically justs shifts the interval one way and then back in the opposite direction the same amount.</p>
<p>Answer is I and II</p>
<p>I and II.</p>
<p>You can do this by actually choosing a curve and testing it. For example, take x^2 on the interval 1-3. You get 26/3. Well, do the property of III. Your intergral becomes (x+1)^2 from 0 to 2. You get 26/3.</p>
<p>Conceptually, it could help if I could draw a graph. But image the graph and the area shaded of x^2 from 1 to 3. Now, since A is 1, you change the graph to (x+1)^2. (When you add a number to x in parentheses, the graph shifts left.) Now, the interval is now 0 to 2. Imagine this (the area shaded from 0 to 2 of x^2 shifted to the left). The area is the same exact thing! It’s just shifted over to the left.</p>
<p>ok that makes sense, it cant be III because c could be anything causing III to not necessarily have the same value as I & II?</p>
<p>…if you just write out the integrals then you’ll see that II is just a shifting of one. To change the bounds of an integral, if you add to the lower and upper bounds, then you have to subtract that value from the x value of the function. (shrug) It’s not that hard of a question. What you really should be worried about is the convergence series. At least for me, that the hardest.</p>
<p>akai, that is a lot easier, guess we never learned that in my calc class. btw im taking AB not BC; lucky me.</p>
<p>oh, my bad.</p>