Though it may not be apparent at first, the study of derivatives (Calculus I) and integrals (Calculus II) are intimately related. Differentiation and integration are inverse operations.

You have previously studied the inverse relationship for functions. Recall that for any bijective function $f$ (a one-to-one relationship) there exists an inverse functions $f^{-1}$ which undoes the effects of $f$: \[ (f^{-1}\!\circ f) (x) \equiv f^{-1}(f(x)) = x. \] and \[ (f \circ f^{-1}) (y) \equiv f(f^{-1}(y)) = y. \] The circle $\circ$ stands for composition of functions, i.e., first you apply one function and then you apply the second function. When you apply a function followed by its inverse to some input you get back the original input.

The integral is the “inverse operation” to the derivative. If perform the integral operation followed by the derivative operation on some function, you will get back the same function. This is stated more formally as the Fundamental Theorem of Calculus.

Let $f(x)$ be a continuous function and let $F(x)$ be its antiderivative on the interval $[a,b]$: \[ F(x) = \int_a^x f(t) \; dt, \] then, the derivative of $F(x)$ is equal to $f(x)$: \[ F'(x) = f(x), \] for any $x \in (a,b)$.

Thus, we see that differentiation is the inverse operation of integration. We obtained $F(x)$ by integrating $f(x)$. If we then take the derivative of $F(x)$ we get back to $f(x)$. It works the other way too. If you integrate a function and then take its derivative, you get back to the original function. Differential calculus and integral calculus are two sides of the same coin. If you understand this fact, then you understand something very deep about calculus.

Note that $F(x)$ is not a unique anti-derivative. We can add an arbitrary constant $C$ to $F(x)$ and it will still satisfy the above conditions since the derivative of a constant is zero.

If you are given some function $f(x)$, you take its integral and then take the derivative of the result, you will get back the same function: \[ \left(\frac{d}{dx} \circ \int dx \right) f(x) \equiv \frac{d}{dx} \int_a^x f(t) dt = f(x). \] Alternately, you can first take the derivative, and then take the integral, and you will get back the function (up to a constant): \[ \left( \int dx \circ \frac{d}{dx}\right) f(x) \equiv \int_a^x f'(t) dt = f(x) - f(a). \]

Note that we had to use a dummy variable $t$ inside the integral since $x$ is used in the limit. Indeed, all integrals are functions of their limits and the inner variable is not important: we could write $\int_a^x f(y)\;dy$ or $\int_a^x f(z)\;dz$ or even $\int_a^x f(\xi)\;d\xi$ and the answer for all of these will be $F(x)-F(a)$.

As a consequence of the Fundamental theorem, you can reuse all your knowledge of differential calculus to solve integrals.

Suppose you are asked find this integral: \[ \int x^2 dx. \] Using the Fundamental theorem, we can rephrase this question as the search for some function $F(x)$ such that \[ F'(x) = x^2. \] Now since you remember your derivative formulas well, you will guess right away that $F(x)$ must contain a $x^3$ term. This is because you get back quadratic term when you take the derivative of cubic term. So we must have $F(x)=cx^3$, for some constant $c$. We must pick the constant that makes this work out: \[ F'(x) = 3cx^2 = x^2, \] therefore $c=\frac{1}{3}$ and the integral is: \[ \int x^2 dx = \frac{1}{3}x^3 + C. \] Did you see what just happened? We were able to take an integral using only derivative formulas and “reverse engineering”. You can check that, indeed, $\frac{d}{dx}\left[\frac{1}{3}x^3\right] = x^2$.

You can also use the Fundamental theorem to check your answers.

Suppose a friend tells you that \[ \int \ln(x) dx = x\ln(x) - x + C, \] but he is a shady character and you don't trust him. How can you check his answer? If you had a smartphone handy, you can check on live.sympy.org, but what if you just have pen and paper? If $x\ln(x) - x$ is really the antiderivative of $\ln(x)$, then by the Fundamental theorem of calculus, if we take the derivative we should get back $\ln(x)$. Let's check: \[ \frac{d}{dx}\!\left[ x\ln(x) - x \right] = \underbrace{\frac{d}{dx}\!\left[x\right]\ln(x)+ x \left[\frac{d}{dx} \ln(x) \right]}_{\text{product rule} } - \frac{d}{dx}\left[ x \right] = 1\ln(x) + x\frac{1}{x} - 1 = \ln(x). \] OK, so your friend is correct.