How to get WolframAlpha to determine $f(2)$ when $f(x)=3x^2-2x+\int_0^2f(t)dt$

Question

This is my first question at StackExchange. It's about WolframAlpha.

The function $f(x)$ is defined as: $$f(x)=3x^2-2x+\int_0^2f(t)dt$$

And I'm trying to find the value of $f(2)$.

So at WolframAlpha, I typed in:

f(x)=3x^2-2x+integral(0,2)f(t)dt ; solve f(2)

and other variations like changing "solve f(2)" to "f(2)=?" but none of these worked.

How can I get this working?

Answer 1

Let $\int_0^2f(t)dt=a$. We get $$f(x)=3x^2-2x+a$$

Let $F'(x)=f(x)$. We get $$F(x)=x^3-x^2+ax+C$$

Therefore: $$\begin{eqnarray*}a&=&\int_0^2f(t)dt=F(2)-F(0)\\&=&(2^3-2^2+2a+C)-(0+C)\Longrightarrow a=-4\end{eqnarray*}$$

So: $$f(x)=3x^2-2x-4\Longrightarrow f(2)=4$$

Answer 2

If you integrate both sides from $0$ to $2$ you get $\int_0^{2}f(t)dt=-4$. Now you know exactly what $f(x)$ is and you can write down $f(2)$.

Answer 3

As long as $\int_0^2 f(t) dt$ is a constant $c$ we proceed with

$$ f(x) = 3x^2-2x + c = 3x^2-2x+\int_0^2(3t^2-2t+c)dt $$

and thus we calculate $c$

$$ c = 4+2c\to c = -4 $$

Answer 4

$f'(x)=6x-2$ get rids of the constant integral, so you can also proceed by integration by parts:

$\displaystyle f(2) =8+\int_0^2f(t)\,dt =8+\bigg[tf(t)\bigg]_0^2-\int_0^2tf'(t)\,dt =8+2f(2)-\int_0^2(6t^2-2t)\,dt =8+2f(2)-12$

Giving $f(2)=4$