Differentiating an integral with variable upper limit

$\int_a^x f(t)\,dt$ changes with its upper limit $x$, so it is a function of $x$. If $F(t)$ is an antiderivative of $f(t)$, it equals $F(x)-F(a)$, and $F(a)$ is a constant that vanishes when differentiated.

For example, $\int_3^x t^2\,dt=\frac{x^3}{3}-9$ differentiates to $x^2$. You get the answer without integrating, just by replacing $t$ with $x$ in $t^2$.

Examples

1. 01This integral has been computed. What do you get when you differentiate it?$\displaystyle \int_{3}^{x} t^{2}\,dt = \frac{x^{3}}{3} - 9$AnswerHide

   Answer$\displaystyle x^{2}$

   Steps

   1. The term from the lower limit is a constant, so it dies
   2. $\displaystyle \frac{d}{dx}\left(\frac{x^{3}}{3} - 9\right)$$\displaystyle = x^{2}$
2. 02Evaluate:$\displaystyle \frac{d}{dx}\int_{0}^{x} t^{3}\,dt$AnswerHide

   Answer$\displaystyle x^{3}$

   Steps

   1. $\displaystyle \int_{0}^{x} f(t)\,dt$$\displaystyle = F(x) - F(0)$
   2. F(0) is a constant, so differentiating kills it
   3. $\displaystyle F'(x)$$\displaystyle = f(x)$
   4. Just replace t with x
   5. $\displaystyle \frac{d}{dx}\int_{0}^{x} t^{3}\,dt$$\displaystyle = x^{3}$
3. 03What do you get when you substitute x=3 into this expression?$\displaystyle \frac{d}{dx}\int_{1}^{x} (t^{2}+1)\,dt$AnswerHide

   Answer$\displaystyle 10$

   Steps

   1. $\displaystyle \int_{1}^{x} f(t)\,dt$$\displaystyle = F(x) - F(1)$
   2. F(1) is a constant, so differentiating kills it
   3. $\displaystyle F'(x)$$\displaystyle = f(x)$
   4. Just replace t with x
   5. $\displaystyle \frac{d}{dx}\int_{1}^{x} (t^{2}+1)\,dt$$\displaystyle = x^{2}+1$
   6. Then put x = 3 in
   7. $\displaystyle 3^{2}+1$$\displaystyle = 10$
4. 04What goes in □?$\displaystyle \frac{d}{dx}\int_{a}^{x} f(t)\,dt = \square$AnswerHide

   Answer$\displaystyle f(x)$

   Steps

   1. $\displaystyle \int_{a}^{x} f(t)\,dt$$\displaystyle = F(x) - F(a)$
   2. F(a) is a constant, so differentiating kills it
   3. $\displaystyle F'(x)$$\displaystyle = f(x)$
5. 05Evaluate:$\displaystyle \frac{d}{dx}\int_{0}^{x} t^{2}\,dt$AnswerHide

   Answer$\displaystyle x^{2}$

   Steps

   1. $\displaystyle \int_{0}^{x} f(t)\,dt$$\displaystyle = F(x) - F(0)$
   2. F(0) is a constant, so differentiating kills it
   3. $\displaystyle F'(x)$$\displaystyle = f(x)$
   4. Just replace t with x
   5. $\displaystyle \frac{d}{dx}\int_{0}^{x} t^{2}\,dt$$\displaystyle = x^{2}$
6. 06Evaluate:$\displaystyle \frac{d}{dx}\int_{1}^{x} (3t+2)\,dt$AnswerHide

   Answer$\displaystyle 3x+2$

   Steps

   1. $\displaystyle \int_{1}^{x} f(t)\,dt$$\displaystyle = F(x) - F(1)$
   2. F(1) is a constant, so differentiating kills it
   3. $\displaystyle F'(x)$$\displaystyle = f(x)$
   4. Just replace t with x
   5. $\displaystyle \frac{d}{dx}\int_{1}^{x} (3t+2)\,dt$$\displaystyle = 3x+2$

FAQ

Q1Does the lower limit $a$ affect the answer?AnswerHide

A

No. Changing $a$ only changes the constant $F(a)$, whose derivative is $0$. Both $\frac{d}{dx}\int_1^x(t^2+1)\,dt$ and $\frac{d}{dx}\int_2^x(t^2+1)\,dt$ equal $x^2+1$.

Q2Is $\int_2^4 t^3\,dt$ also a function of $x$?AnswerHide

A

No. Both limits are numbers, so its value is a constant ($60$) that does not depend on $x$. It is a function of $x$ only when $x$ appears in a limit.