Question:

Consider the function $f(x) = \int_{1}^{x} \frac{2t^2+3}{t+1} \,dt$, where $x \geq 1$.

a) Find $f'(x)$ using the Fundamental Theorem of Calculus.

b) Using the result from part (a), evaluate $f'(2)$.

c) Consider the function $g(x) = \int_{0}^{x} f(t) \,dt$. Find an expression for $g'(x)$.

d) Using the result from part (c), evaluate $g'(2)$.

Answer:

a) The Fundamental Theorem of Calculus states that if $f(x) = \int_{a}^{x} F(t) \,dt$, then $f'(x) = F(x)$. In this case, we have $f(x) = \int_{1}^{x} \frac{2t^2+3}{t+1} \,dt$.

To find $f'(x)$, we use the Chain Rule. Let $u = x+1$ and $g(u) = \int \frac{2(u-1)^2+3}{u} \,du$. Applying the Chain Rule, we have:

$\frac{d}{dx} \left[ \int_{1}^{x} \frac{2t^2+3}{t+1} \,dt \right] = \frac{d}{dx} \left[ g(x+1) \right]$

$= g'(x+1) \cdot \frac{d}{dx} \left[ x+1 \right]$

$= g'(x+1)$

Now, to find $g(u)$, we integrate $\frac{2(u-1)^2+3}{u}$ with respect to $u$. First, we expand and simplify the integrand:

$g(u) = \int \frac{2(u^2-2u+1)+3}{u} \,du$

$= \int \frac{2u^2-4u+2+3}{u} \,du$

$= \int \left(2u-4 + \frac{2+3}{u}\right) \,du$

$= \int \left(2u-4 + \frac{5}{u}\right) \,du$

Now we integrate term by term:

$= 2 \int u \,du - 4 \int \,du + 5\int \frac{1}{u} \,du$

$= u^2 - 4u + 5 \ln|u| + C$

Finally, substituting $u = x+1$, we get:

$g(x+1) = (x+1)^2 - 4(x+1) + 5 \ln|x+1| + C$

Therefore, $f'(x) = g'(x+1) = \frac{d}{dx} \left[ (x+1)^2 - 4(x+1) + 5 \ln|x+1| + C \right]$

b) To evaluate $f'(2)$, we substitute $x = 2$ into the derivative expression we found in part (a):

$f'(2) = \frac{d}{dx} \left[ (x+1)^2 - 4(x+1) + 5 \ln|x+1| + C \right] \bigg|_{x=2}$

$= \frac{d}{dx} \left[ (2+1)^2 - 4(2+1) + 5 \ln|2+1| + C \right]$

$= \frac{d}{dx} \left[ 9 - 12 + 5 \ln|3| + C \right]$

$= \frac{d}{dx} \left[ -3 + 5 \ln|3| + C \right]$

$= 0$

Therefore, $f'(2) = 0$.

c) To find $g'(x)$, we apply the Fundamental Theorem of Calculus again. We have $g(x) = \int_{0}^{x} f(t) \,dt$.

Applying the Fundamental Theorem of Calculus, we get:

$g'(x) = f(x)$

$= \int_{1}^{x} \frac{2t^2+3}{t+1} \,dt$

d) To evaluate $g'(2)$, we can substitute $x = 2$ into the integrand expression we found in part (c):

$g'(2) = \int_{1}^{2} \frac{2t^2+3}{t+1} \,dt$

$= \left[ \int_{1}^{2} \frac{2t^2+3}{t+1} \,dt \right]$

$= \left[ \int \frac{2t^2+3}{t+1} \,dt \right] \bigg|_{1}^{2}$

$= \left[ 2t - 4 \ln|t+1| \right] \bigg|_{1}^{2}$

$= \left[ 2(2) - 4 \ln|2+1| \right] - \left[ 2(1) - 4 \ln|1+1| \right]$

$= 4 - 4 \ln(3) - 2 + 4 \ln(2)$

$= 2 + 4 \ln(2) - 4 \ln(3)$

Therefore, $g'(2) = 2 + 4 \ln(2) - 4 \ln(3)$.