Substitution in definite integrals

After substituting, find the values of $t$ that correspond to the endpoints in $x$, and use them as the new limits.

Here $t=2x+1$ gives $dt=2\,dx$, so $dx=\frac{1}{2}\,dt$; $x=0$ gives $t=1$ and $x=1$ gives $t=3$. Once the limits are changed, there is no need to go back to $x$ at the end.

Examples

1. 01What is t when x=1?$\displaystyle \begin{gathered} \displaystyle \int_{0}^{1} (2x+2)^{3}dx, \\[6pt] \displaystyle t=2x+2 \end{gathered}$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. Put the upper limit into the same expression
   2. $\displaystyle t$$\displaystyle = 2x+2$$\displaystyle x=1$$\displaystyle \Rightarrow t = 4$
   3. Do not leave the limit as an $x$ value; forgetting this is the trap of this atom
2. 02What is the value?$\displaystyle \frac{1}{2}\int_{1}^{3} t^{3}\,dt$AnswerHide

   Answer$\displaystyle 10$

   Steps

   1. $t^{3}$ integrates to $\frac{t^{4}}{4}$; subtract the lower end from the upper, then apply the $\frac{1}{2}$ in front
   2. $\displaystyle \frac{1}{2}\cdot\frac{3^{4}-1^{4}}{4}$$\displaystyle = 10$
   3. Forgetting that $\frac{1}{2}$ gives 20 instead
3. 03What goes in □?$\displaystyle t=3x+2 \;\Rightarrow\; dx = \square\,dt$AnswerHide

   Answer$\displaystyle \frac{1}{3}$

   Steps

   1. Differentiate $t$ with respect to $x$
   2. $\displaystyle t$$\displaystyle = 3x+2$$\displaystyle \frac{dt}{dx}$$\displaystyle = 3$
   3. So $dt = 3\,dx$; to get $dx$ you divide by $3$, making $\square$ equal to $\frac{1}{3}$
4. 04What is t when x=0?$\displaystyle \begin{gathered} \displaystyle \int_{0}^{2} (3x+2)^{3}dx, \\[6pt] \displaystyle t=3x+2 \end{gathered}$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. Put the lower limit straight into the expression you set
   2. $\displaystyle t$$\displaystyle = 3x+2$$\displaystyle x=0$$\displaystyle \Rightarrow t = 2$
   3. With $x=0$ the $3x$ term drops out, leaving just $2$
5. 05What is t when x=1?$\displaystyle \begin{gathered} \displaystyle \int_{0}^{1} (2x+3)^{3}dx, \\[6pt] \displaystyle t=2x+3 \end{gathered}$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. Put the upper limit into the same expression
   2. $\displaystyle t$$\displaystyle = 2x+3$$\displaystyle x=1$$\displaystyle \Rightarrow t = 5$
   3. Do not leave the limit as an $x$ value; forgetting this is the trap of this atom
6. 06What is the value?$\displaystyle \frac{1}{2}\int_{1}^{5} t^{3}\,dt$AnswerHide

   Answer$\displaystyle 78$

   Steps

   1. $t^{3}$ integrates to $\frac{t^{4}}{4}$; subtract the lower end from the upper, then apply the $\frac{1}{2}$ in front
   2. $\displaystyle \frac{1}{2}\cdot\frac{5^{4}-1^{4}}{4}$$\displaystyle = 78$
   3. Forgetting that $\frac{1}{2}$ gives 156 instead

FAQ

Q1Do I need to switch back to $x$ before substituting the limits?AnswerHide

A

No. If the limits have been changed to values of $t$, you can finish the calculation in $t$. You switch back to $x$ only if you kept the limits in $x$.

Q2If the new lower limit is larger than the upper one, should I swap them?AnswerHide

A

No. Write the value that corresponds to the lower limit in $x$ as the lower limit, and the one for the upper limit as the upper limit, as they are.