Question

q6 (1 point)
if $f(x) = \int_{1}^{x} \ln t^{2}dt$, then $f(2) =$
$\ln(2^{2})$
$\int_{1}^{2} \ln t^{2}dt$
$0$
$\ln 2^{2} - \ln 1^{2}$
$\int_{1}^{2} \ln 2^{2}dt$

Explanation:

Step1: Apply Fundamental Theorem of Calculus

By the First Fundamental Theorem of Calculus, if $f(x) = \int_{a}^{x} g(t)dt$, then $f'(x) = g(x)$. Here, $g(t)=\ln t^2$, so $f'(x)=\ln x^2$.

Step2: Substitute $x=2$ into $f'(x)$

Replace $x$ with 2 in the derivative function.
<Expression>
$f'(2)=\ln(2^2)$
</Expression>

Answer:

$\ln(2^2)$