Antiderivatives

What Nour says

1. 1. What is an antiderivative?

   Today, the whole chapter: "Antiderivatives", Year 4, Mathematics section.

   We will build this step by step, from the idea to full worked problems.

   By the end, you can find any antiderivative that fits a given condition.

   Let's start with the first idea: what is an antiderivative.

   Suppose we know a function f, and we look for another function F.

   We want F whose derivative gives back exactly f.

   This F is called an antiderivative of f on an interval.

   In other words, F is a function whose derivative is f itself.

   So finding an antiderivative is the reverse of finding a derivative.

   Let's see why this idea makes sense, with a simple picture.

   Take f of x equal to 2x, a simple function we already know well.

   We ask: which function F has derivative equal to 2x everywhere.

   We remember that the derivative of x squared is exactly 2x.

   So F of x equal to x squared is one antiderivative of f.

   But notice something important: x squared plus 1 also works.

   Its derivative is still 2x, because the derivative of a constant is zero.

   So there is never just one antiderivative: there are infinitely many.

   All these curves are vertical shifts of each other, by a constant.

   Now let's work through a full example, step by step.

   We look for F with F prime of x equal to 3x squared.

   We recall that the derivative of x cubed is 3x squared.

   So F of x equal to x cubed is a valid antiderivative.

   We check it by taking the derivative of x cubed again.

   The derivative of x cubed is indeed 3x squared, which is f.

   So our answer is correct: F of x equal to x cubed works.

   We could also write F of x equal to x cubed plus 5.

   Adding any constant still gives derivative equal to 3x squared.

   So the full family of antiderivatives is x cubed plus any constant k.

   Let's state the key result clearly now.

   If F is one antiderivative of f, every other one is F plus a constant.

   This whole family of functions is what we study in this chapter.

   If F is an antiderivative of f, what is F(x) + 7 also?

2. 2. Antiderivatives of a continuous function

   Now let's move to antiderivatives of a continuous function.

   We already know continuity from an earlier chapter of this year.

   A continuous function has no jump and no break in its curve.

   Here is the key theorem of this section, stated simply.

   Every continuous function on an interval has at least one antiderivative.

   This is why we call them primitives of continuous functions in the manual.

   That is, continuity guarantees we can always find such an F.

   Let's see why this matters, with a limit case.

   If f had a break, like a jump at one point, things go wrong.

   No smooth function F can have this jumpy function as derivative.

   So continuity is the condition that makes the theorem work.

   Now let's apply this with a full worked example.

   f of x equals x squared plus 1 is continuous on all real numbers.

   So by the theorem, f has an antiderivative F on all of ℝ.

   We check by differentiating: derivative of x cubed over 3 is x squared.

   And the derivative of x is just 1.

   Adding both parts, F prime of x equals x squared plus 1.

   So F is indeed an antiderivative of f, confirmed by the derivative.

   Notice we chose one F among infinitely many, by picking a constant.

   We could write F plus 3, or F minus 7, and it still works.

   Let's close with the key result of this section.

   Continuity alone is enough to guarantee an antiderivative exists.

   The function does not need any other special property.

   This theorem is the foundation for everything we do next.

   Does every continuous function on an interval have an antiderivative there?

3. 3. Antiderivatives of usual functions

   Now let's learn how to calculate antiderivatives of usual functions.

   That is, we build a table of standard results we can use directly.

   Let's start with constant functions, the simplest case.

   An antiderivative of a constant k is F of x equal to kx.

   Next, the power functions, which appear everywhere in this chapter.

   An antiderivative is x to the power n plus 1, over n plus 1.

   We check: the derivative of x to n plus 1 brings down n plus 1.

   That factor cancels with the denominator, leaving exactly x to the n.

   Next, the function 1 over x squared, for x different from 0.

   Its antiderivative is minus 1 over x.

   Next, the function 1 over the square root of x, for x greater than 0.

   Its antiderivative is 2 times the square root of x.

   We remember these from differentiation, just read in reverse direction.

   Let's see why this reverse reading works, using one case.

   The derivative of square root of x is 1 over 2 square root of x.

   So 2 times square root of x has derivative 1 over square root of x.

   This confirms our table entry for the square root case.

   Now let's apply the whole table in one full example.

   We split f into two simple pieces: x cubed, and 1 over x squared.

   For x cubed, we use n equals 3 in our power rule.

   For 1 over x squared, our table gives minus 1 over x.

   Putting both pieces together gives our full antiderivative F.

   Let's check by differentiating each term of F again.

   And the derivative of minus 1 over x gives 1 over x squared.

   Adding both derivatives gives back exactly our original function f.

   So our table of usual functions lets us solve many problems fast.

   Each formula comes directly from a derivative we already know.

   Memorizing this short table is the key tool for this whole chapter.

   What is an antiderivative of $f(x) = x^5$?

4. 4. Operations on antiderivatives

   Now let's learn the operations on antiderivatives.

   That is, rules that let us handle sums and multiples of functions.

   These properties and operations on antiderivatives save us a lot of work.

   First rule: the antiderivative of a sum is the sum of antiderivatives.

   Here F is an antiderivative of f, and G is an antiderivative of g.

   Second rule: a constant factor just stays in front, unchanged.

   Let's see why this makes sense, using the derivative rules we know.

   The derivative of F plus G is F prime plus G prime.

   So F plus G really is an antiderivative of f plus g.

   Same idea for the constant: derivative of k times F is k times f.

   These two rules let us break any function into simple pieces.

   Now let's use both rules in one full worked example.

   We treat each term separately, using our table from before.

   An antiderivative of x squared is x cubed over 3.

   An antiderivative of x is x squared over 2.

   An antiderivative of the constant 4 is 4x.

   Adding all three pieces gives our complete antiderivative F.

   Let's check this by differentiating F term by term.

   The second term gives minus 3x, and the third gives 4.

   Adding these back gives exactly 5x squared minus 3x plus 4.

   So our rules for sums and constants worked perfectly on this example.

   We never need to guess: we just apply the table piece by piece.

   Let's state the key result of this section clearly.

   The antiderivative of a sum splits into separate antiderivatives.

   And constant factors always stay outside, untouched by the operation.

   What is an antiderivative of $f(x) = 6x^2 + 2$?

5. 5. Finding the antiderivative with one condition

   Now the final idea: finding the antiderivative with one condition.

   We call this determining the set of antiderivatives, then picking just one.

   That is, among all antiderivatives, we use one extra fact to choose exactly one.

   We know that f has infinitely many antiderivatives, each differing by a constant.

   One extra condition, like F of some point, fixes that constant exactly.

   This turns infinitely many functions into just one single answer.

   Let's see why one condition is always enough here.

   The family of antiderivatives is a set of parallel shifted curves.

   Fixing one point picks exactly one curve from that whole family.

   Now let's solve a full example, step by step.

   We first find the general antiderivative of f, using our table.

   Here x squared comes from 2x, and x comes from the constant 1.

   Now we use the condition: F of 1 must equal 4.

   This gives 2 plus k equals 4.

   Solving for k, we get k equal to 2.

   So the one antiderivative that fits our condition is found.

   Let's check: F of 1 equals 1 plus 1 plus 2, which is 4.

   That matches the condition F of 1 equal to 4 exactly.

   So the condition fixed one unique function from the whole family.

   Let's state the key method for this type of problem.

   First, find the general antiderivative with a constant k inside it.

   Then use the given condition to compute the exact value of k.

   If $F(x) = x^3 + k$ and $F(0) = 5$, what is $k$?

   That completes the chapter on antiderivatives, from the basic idea to full problem solving.

   You now know the table of usual functions, the key operations, and how one condition fixes a unique answer.

   Practice both exercises carefully, and check every answer by differentiating it back.