# Solving integral without fundamental theorem of calculus

Solve $\int_0^1 3xdx$ without using the fundamental theorem of calculus.

I know that, to solve an integral without the fundamental theorem of calculus, I can find the upper sum and the lower sum. I can write down these sums. However, since they have several variables, how do I obtain just one integral value from them?

• What do you mean several variables? – Zhoe Feb 9 '14 at 0:05
• For example upper sum $U_p(3x)= \sum_{i=1}^k \frac{3i}{k}$ has i and k – kiwifruit Feb 9 '14 at 0:18

Following your comment, the upper sum of a partition where every subinterval is of equal witdh is $\displaystyle \sum_{i=1}^k \frac{3i}{k^2} = \frac{3}{k^2}\sum_{i=1}^k i = \frac{3}{k^2} \frac{k (k+1)}{2} = \frac{3}{2}\frac{k+1}{k}$, where $k$ is the number of subintervals. As $k$ goes to $+\infty$, $\displaystyle \frac{k+1}{k} \to 1$ and thus $\displaystyle \frac{3}{2}\frac{k+1}{k} \to \frac{3}{2}$. Therefore the infimum of the upper sums is $\displaystyle \frac{3}{2}$.

• This is the solution I am looking for, but where did the $\frac{k(k+1)}{2}$ part come from? – kiwifruit Feb 9 '14 at 1:03
• @kiwifruit See en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF – dani_s Feb 9 '14 at 1:06
• @kiwifruit There was actually a mistake, each rectangle has base $\frac{1}{k}$ and height $\frac{3i}{k}$. Therefore the area of each rectangle is $\frac{3i}{k^2}$. And of course you actually need $k$ to go to $+\infty$. I have corrected my answer. – dani_s Feb 9 '14 at 1:27

$\int_0^1 3 x dx$ is the area of the triangle of base $1$ and height $3$. So the answer is $3/2$

This result uses just the property if integrals as the area under the curve.

• So it's ok to use the full area without partitioning? – kiwifruit Feb 9 '14 at 0:21
• yes as long as you use the fact that the area of a triangle is half base times height. Looks like this is what is expected. – user44197 Feb 9 '14 at 0:23
• It is perfectly acceptable to use formulas from classical geometry to evaluate definite integrals where possible (you will find it needful for a while when confronted with integrals such as $\ \int_{-1}^1 \ \sqrt{1-x^2} \ \ dx \ ,$ which you won't be able to solve directly even using a $\ u-$ substitution). Calculus was invented in part to compute the areas of closed curves and surface areas and volumes of solids for shapes which are beyond the "reach" of classical geometry. – colormegone Feb 9 '14 at 1:02