Question

Find the values of c that satisfy MVT for integrals on \[\left[ {0,1} \right]\] where, $f(x) = x(1 - x)$ .
A.\[c = \dfrac{1}{2}\]
B.\[c = - \dfrac{1}{2}\]
C.\[c = \dfrac{2}{3}\]
D.\[c = \dfrac{1}{3}\]

Answer 1

Hint: We will find the value of c by applying MVT on the function as given in the question. The formula of MVT is given by MVT ${f^{'}}(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$ ; where a and b are the lower and upper limits of the given function and \[f\left( a \right)\] and \[f\left( b \right)\] will be obtained by placing \[x{\text{ }} = {\text{ }}a\] and \[x{\text{ }} = {\text{ }}b\] in the given equation.

Complete step-by-step answer:
Here the given function is continuous and differentiable in [0,1]
Hence, a=0 and b=1
$f(x) = x(1 - x)$
If we put a=0 and b=1 in the given function we get
f(a)=0(1-0)=0
f(b)=1(1-1)=0
So f(a)=f(b)=0
Now by applying MVT in the given function ,
$ \Rightarrow {f^{'}}(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
Putting value of f(a),f(b),a and b
$\Rightarrow {f^{'}}(c) = \dfrac{{0 - 0}}{{0- 0}}$
After differentiating the given function
$
   \Rightarrow (1 - c) - c = 0 \\
   \Rightarrow 1 - 2c = 0 \\
   \Rightarrow c = \dfrac{1}{2} \\
 $
Therefore the value of \[c = \dfrac{1}{2}\] .

Note: Mean Value Theorem (MVT) is defined as The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if \[f(x)\] is defined and continuous on the interval \[[a,b]\] and differentiable on\[(a,b)\], then there is at least one number c in the interval \[(a,b)\] i.e. \[a\; < \;c\; < \;b\] such that ${f^{'}}(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$