AP Calculus AB Exam Question:

Consider the function $f(x) = \frac{3}{x^2+1}$. Let $F(x)$ be the indefinite integral of $f(x)$ and let $A$ be the definite integral of $f(x)$ on the interval $[0, 2]$.

a) Find $F(x)$.

b) Evaluate the value of $A$.

c) Find the average value of $f(x)$ on the interval $[0, 2]$.

Answer:

a) To find $F(x)$, we need to find the indefinite integral of $f(x)$. Recall that the integral of $\frac{1}{x^2+1}$ can be found using the inverse tangent function, $\tan^{-1}(x)$.

We can rewrite $f(x)$ as $f(x) = \frac{3}{1+x^2}$.

So, $F(x) = \int f(x) \, dx$

$F(x) = \int \frac{3}{1+x^2} \, dx$

Using the substitution $u = 1+x^2$, we have $du = 2x \, dx$. Rearranging, we get $dx = \frac{1}{2x} \, du$.

$F(x) = \int \frac{3}{u} \cdot \frac{1}{2x} \, du$

$F(x) = \frac{3}{2} \int \frac{1}{u} \, du$

$F(x) = \frac{3}{2} \ln|u| + C$

Substituting back $u = 1+x^2$:

b) To evaluate the definite integral $A = \int_0^2 f(x) \, dx$, we can directly substitute the limits into $F(x)$ and subtract:

c) The average value of a function $f(x)$ on an interval $[a, b]$ is given by:

In this case, we want to find the average value of $f(x)$ on the interval $[0, 2]$:

Using the same substitution as before, $u = 1+x^2$, we have:

Thus, the average value of $f(x)$ on the interval $[0, 2]$ is also $\frac{3}{2} \ln 5$.