# Suppose $\int_0^9f(t)dt=12$. Then is it true that $\int_0^3f(3x)dx = 12$

Suppose $\int_0^9f(t)dt=12$. Then is it true that $\int_0^3f(3x)dx = 12$ ?

How do I go about figuring this out? I tried differentiating and using fundamental theorem of calculus but couldn't figure it out. Please help.

• Try the substitution $t=3x$ and see what happens. – Ethan Bolker Apr 26 '16 at 15:52
• Geometrically, you've just shrunk the graph in the $x$ direction without changing the $y$ direction, so how could the area stay the same? If $f(x)=4/3$ for all $x$, for example, your original is the are of a rectangle with width $9$ and height $4/3$, for total area $12$. But $\int_0^{3}f(3x)dx$ represents the area of a rectangle with width $3$ and height $4/3$ for area $4$. – Thomas Andrews Apr 26 '16 at 15:56
• @EthanBolker I was able to understand it geometrically. But I want to understand where you were leading me. Can you please give an elaborate answer? – Anirudh Gangwal Apr 26 '16 at 15:59
• @ThomasAndrews has answered your question. – Ethan Bolker Apr 26 '16 at 16:06

Geometrically, you've just shrunk the graph in the $x$ direction without changing the $y$ direction, so how could the area stay the same?
If $f(t)=4/3$ for all $t$, for example, your original is the are of a rectangle with width $9$ and height $4/3$, for total area $12$. But $\int_0^{3}f(3x)dx$ represents the area of a rectangle with width $3$ and height $4/3$ for area $4$.
If $f(t)=\frac{8}{27}t$, then $\int_{0}^{9} f(t)\,dt$ represents a triangle with base $9$ and height $\frac{8}{17}\cdot 9 = \frac{8}{3}$, so the area is $\frac{1}{2}8\cdot\frac{8}3=12$ But if you draw the graph of $f(3x)$ for $x\in [0,3]$ you see that you get a triangle with the same height, but base $3$, so the area is $4$.
In fact, you always get $4$. This is, geometrically, the fact that scaling a figure in one direction by a factor of $c$ multiplies the area by of that figure by $c$. In this case, $c=\frac{1}{3}$.
Hint: Try the change of variable $t=3x.$