# Integrals with variables on top and bottom

How do I solve an integral that has variables on both top and bottom? To solve an integral like $\int^{x}_0$ $t^2$+5 dt, I would simply plug in x for t, and if both the top and bottom of an integral are constants, I would subtract what the answer is at the bottom constant from the answer at the top constant. But how would I solve an integral like $\int^{x^2}_x$ $t^2$dt? If I knew how to solve it, then I might be able to take the derivative of whatever it is, maybe even the second derivative. How do I get the derivative of $\int^{x^2}_x$ $t^2$dt?

• What do you mean, "it didn't work"? Apr 28, 2013 at 1:52

Well, direct substitution of the result applies as usual. But if it helps, think of it as an integral from the lower bound to some point between them, and then an integral from that point to the upper bound. That is,

$$\int_x^{x^2} t^2 dt = \int_x^a t^2 dt + \int_a^{x^2} t^2 dt$$ Depending on the function, you may also be able to use a point outside of the range of the variables, and use this: $$\int_x^{x^2} t^2 dt = \int_a^{x^2} t^2 dt - \int_a^x t^2 dt$$ $a=0$ is an obvious special case of this, and helps you to see what is going on. But it requires that the two integrals on the right be finite, which is not necessarily true for any function (for instance, if the function is $\frac1t$, the integrals will diverge).

So, in the case of the function you provided, we have three ways of thinking of it: $$\int_x^{x^2} t^2 dt = \left[\frac{t^3}3\right]_x^{x^2} = \frac{(x^2)^3}3-\frac{x^3}3\\ \int_x^{x^2} t^2 dt = \int_x^a t^2 dt + \int_a^{x^2} t^2 dt=\left[\frac{t^3}3\right]_x^a+\left[\frac{t^3}3\right]_a^{x^2} = \frac{a^3}3-\frac{x^3}3+\frac{(x^2)^3}3-\frac{a^3}3=\frac{(x^2)^3}3-\frac{x^3}3\\ \int_x^{x^2} t^2 dt = \int_0^{x^2} t^2 dt - \int_0^x t^2 dt = \left[\frac{t^3}3\right]_0^{x^2}+\left[\frac{t^3}3\right]_0^x = \frac{(x^2)^3}3-0+0-\frac{x^3}3=\frac{(x^2)^3}3-\frac{x^3}3$$ Same result each time.

If $F'=f$, then $\int_r^s f=F(s)-F(r)$, and it doesn't matter whether $r$ and $s$ are constants or functions.

Also, to take the derivative of the integral, you don't need to explicitly solve it. You can use the Leibniz Rule, which says: $$\frac{d}{dx}\int_{f_1(x)}^{f_2(x)}g(t)dt = g(f_2(x))\times f_2'(x) - g(f_1(x))\times f_1'(x)$$

So here, $$\frac{d}{dx}\int_{x}^{x^2}t^2dt = (x^2)^2\times (2x)- (x)^2\times (1) =2x^5-x^2$$

You can also get the value of the original integral by integrating this expression, which may be easier.