The accurate choices are (A), (B), and (C). Given that f′′(x)>0 and f(x)−x=0, we need to determine the number of solutions. Let g(x)=f(x)−x, which implies g′(x)=f′(x)−1 and g′′(x)=f′′(x)>0. This indicates the presence of concave possibilities
The correct options are (A),(B) and (C).
Evaluate: \[ \int_1^5 \left( |x-2| + |x-4| \right) \, dx \]
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C