Finding a function defined with a definite integral

$\int_0^3 f(t)\,dt$ contains no $x$; it is a constant. Calling it $k$, the equation $f(x)=2x+\int_0^3 f(t)\,dt$ becomes $f(x)=2x+k$. Integrating this $f$ over $[0,3]$ must give $k$ back.

Solving gives $k=-\frac{9}{2}$, so $f(x)=2x-\frac{9}{2}$. The steps are: set the integral equal to $k$, write $f(x)$ in terms of $k$, and integrate that expression to get an equation for $k$.

Examples

1. 01What is f(x)?$\displaystyle f(x)=6x+k, \quad k=-12$AnswerHide

   Answer$\displaystyle 6x-12$

   Steps

   1. Put k in
   2. $\displaystyle f(x)$$\displaystyle = 6x-12$
   3. Integrating this f gives k back
   4. $\displaystyle \int_{0}^{2} (6x-12)\,dx$$\displaystyle = -12$
2. 02What do you get by integrating this over [0, 3]?$\displaystyle f(x)=2x+k$AnswerHide

   Answer$\displaystyle 9+3k$

   Steps

   1. $\displaystyle \int_{0}^{3} 2x\,dx$$\displaystyle = 9$
   2. k is integrated too, so it picks up the interval length
   3. $\displaystyle \int_{0}^{3} k\,dx$$\displaystyle = 3k$
3. 03Write f(x) in terms of k.$\displaystyle f(x) = 6x + \int_{0}^{3} f(t)\,dt,\; \int_{0}^{3} f(t)\,dt = k$AnswerHide

   Answer$\displaystyle 6x+k$

   Steps

   1. The integral part is a number, so call it k
   2. $\displaystyle f(x)$$\displaystyle = 6x+k$
   3. k is a number: it is not multiplied by x
4. 04Find k.$\displaystyle k = 48+4k$AnswerHide

   Answer$\displaystyle -16$

   Steps

   1. $\displaystyle k = 48+4k$$\displaystyle \Rightarrow -3k = 48$
   2. Divide both sides by -3
   3. $\displaystyle k$$\displaystyle = -16$
   4. Put it back and it checks out
   5. $\displaystyle 48 + 4 \cdot (-16)$$\displaystyle = -16$
5. 05Write f(x) in terms of k.$\displaystyle f(x) = 2x + \int_{0}^{3} f(t)\,dt,\; \int_{0}^{3} f(t)\,dt = k$AnswerHide

   Answer$\displaystyle 2x+k$

   Steps

   1. The integral part is a number, so call it k
   2. $\displaystyle f(x)$$\displaystyle = 2x+k$
   3. k is a number: it is not multiplied by x
6. 06What is f(x)?$\displaystyle f(x)=4x+k, \quad k=-8$AnswerHide

   Answer$\displaystyle 4x-8$

   Steps

   1. Put k in
   2. $\displaystyle f(x)$$\displaystyle = 4x-8$
   3. Integrating this f gives k back
   4. $\displaystyle \int_{0}^{2} (4x-8)\,dx$$\displaystyle = -8$

FAQ

Q1Why can $\int_a^b f(t)\,dt$ be treated as a constant?AnswerHide

A

Both limits are numbers, so no $x$ remains after integrating and the result is a single number. An integral with $x$ as a limit, such as $\int_a^x f(t)\,dt$, is a function of $x$ and cannot be set equal to a constant.

Q2Why is the variable of integration $t$?AnswerHide

A

To keep it separate from the $x$ in $f(x)$. The value of a definite integral does not depend on the letter used, so $\int_0^3 f(t)\,dt=\int_0^3 f(x)\,dx$.