Formula

$${\int }_0^x\left[{\int }_0^u\phi (t)dt\right]du={\int }_0^x\left[{\int }_t^x\phi (t)du\right]dt$$

Explanation

As for double integrals

Maybe we can understand the double integrals from the perspective of code, and the above formula are equivalent to the following two sum.

sum1 = 0
for u in 0:d1:x:
for t in 0:d2:u:
sum1 += varphi(t)*d1*d2
print sum1
sum2 = 0
for t in 0:d1:x:
for u in t:d2:x:
sum2 += varphi(t)*d1*d2
print sum2

First code:
u=0 at first, and after

	for t in 0:d2:u:
sum1 += varphi(t)*d1*d2

sum1 = 0;
Then u become a little bigger, we do this cycle again. Finally, we can get the the volume accumulated by $\phi (t)$ on this triangle.

Second code:
And yet, we can also treat this idea from another angle. In the first place, t=0, and we execute：

	for u in t:d2:x:
sum2 += varphi(t)*d1*d2

After t increase, we do this cycle again and again, we can also get the the volume accumulated by $\phi (t)$ on this triangle.

That’s all.
#Purpose
How does this artifice work in reality?

I don’t know if you notice that or not, but

sum2 = 0
for t in 0:d1:x :
for u in t:d2:x :
sum2 += varphi(t)*d1*d2
print sum2

is equivalent to

sum2 = 0
for t in 0:d1:x :
sum2 += varphi(t)*d1*(x-t)
print sum2

The reduction in the number of dimensions reduces the number of cycles. That’s why it’s useful.

$${\int }_0^x\left[{\int }_0^u\phi (t)dt\right]du={\int }_0^x\left[{\int }_t^x\phi (t)du\right]dt={\int }_0^x\left[(x-t)\phi (t)\right]dt$$

最后

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