Question

Find the values of $c$ that satisfy the MVT for integrals on $[-2,3]$.
$f\left(t\right)=8t+e^{-3t}$

• A. $c=-1.0973$
• B. $c=-0.0973$
• C. $c=1.0973$
• D. $c=0.0973$

Answer 1

The correct option is C $c=1.0973$

MVT means there exists some $c$ such that $f^1\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given $f\left(t\right)=8t+e^{-3t}$

$\Rightarrow f^1\left(t\right)=8-3e^{-3t}$

From MVT , we get $8-3e^{-3c}=\frac{8(3+2)+e^{-9}-e^6}{5}=\frac{40+e^{-9}-e^6}{5}$

$\Rightarrow 15e^{-3c}=e^6-e^{-9}$

$\Rightarrow c=1.0973$

Therefore correct option is $C$