- #1
brotherbobby
- 618
- 152
- Homework Statement
- For the graph of the function ##f(x)## given below, suppose ##\large{g(x) = \int f(x) \mathrm{d}x}##. Select which of the following are true :##\\[10pt]##
1. ##g(x)## is always positive.
2. ##g(x)## is negative at ##C##.
3. The slope of ##g(x)## at ##B## is zero.
4. ##g(x)## is greater at ##E## than at ##B##.
5. ##g(x)## is smaller at ##C## than at ##F##.
6. ##g(x)## is increasing at ##F##.
- Relevant Equations
- 1. The integral ##g(x)## of a function ##f(x)## is its antiderivative : If ##g(x) = \int f(x) dx\Rightarrow \dfrac{d}{dx} g(x) = f(x)##.
2. The slope of the integral at a point is, therefore, equal to the derivative of the integrand at that point.
3. The amount of an integral between two points ##a## and ##b## on its graph is equal to the area enclosed by that section of the graph and the ##x## axis.
4. A function ##f(x)## is said to be increasing at a point ##x_0## if ##f'(x_0)>0##
Attempt : I argue for or against each statement by putting it down first in blue and my answer in red.
- ##g(x)## is always positive : The exact value of ##g(x)## cannot be determined because an indefinite integral is undetermined upto a constant term ##c## : ##\int f(x) dx = g(x)+c##. Hence, the statement in blue is false. ##\text{ }\large[##However, if ##c## is given to be 0, then we can see that the area enclosed by the graph (curve) ##f(x)## is positive from ##O\rightarrow F## (throughout). This is because a greater (positive) area is enclosed by the graph in the region ##OAB## and ##DEF## than the smaller (negative) area it encloses in the region ##BCD##. However, since nothing is mentioned about the undetermined constant ##c##, no definite value of ##g## can be set against any of the designated points on the graph##\large ]##.##\\[10pt]##
- ##g(x)## is negative at ##C## : This statement is false for the same reason as (1) above. ##\\[10pt]##
- The slope of ##g(x)## at ##B## is zero : The slope of ##g(x)## is the drawn function ##f(x)##. Since ##f(B) = 0##, it implies ##\left.\frac{dg}{dx}\right|_{B} = 0##. The statement is true.##\\[10pt]##
- ##g(x)## is greater at ##E## than at ##B## : Despite the undermined constant ##c## in point 1 above, which leaves the value of ##g(x)## for all points unknown, it is possible to compare the values of ##g(x)## for a pair of given points because this constant term vanishes on taking differences. Calling the origin as ##O##, we find that the area under the graph ##f(x)## from ##O\rightarrow E## is greater that the area under the graph from ##O\rightarrow B##. The integral (value) being an area, the functional value of ##g(E)>g(B)##, implying that the statement above is true. ##\\[10pt]##.
- ##g(x)## is smaller at ##C## than at ##F## : Let us assume, going by "looks" of the regions, we have the ##\text{Area of DEF = ABC > BCD}##. Both at ##C## and at ##F##, the area of region ##OAB## comes in, which can be removed when taking differences. The area of the function at ##F## includes the entire (negative) contribution from ##BCD##, but that is more than compensated by the area of ##DEF## so that ##g(F)>g(C)##. Hence the statement above is true. ##\\[10pt]##
- ##g(x)## is increasing at ##F## : The increasing (or decreasing) behaviour of a function is given by the sign of its derivative, as is well known. Here ##g(x)## is the function and ##f(x)## its derivative. We find from the graph above that ##f(F)>0##. Hence ##g(x)## must be an increasing function at that point, making the statement true. ##\text{ }## [I must add for reasons of clarity that ##f(x)## it itself a decreasing function at ##F## as ##f'(F)<0##. But the statement in question is about its integral ##g(x)##, not the function ##f(x)## itself.
I would like some hints or suggestions if possible, specially if I am mistaken in my thinking above.