Antiderivatives Problems - Calculus - Lecture Notes | MAT 021B, Study notes of Calculus

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4.8 Antiderivatives

Derivative problem Given a function F (x), how to find its derivative f (x)?

• When F (x) is nice enough, we have learned many formulas and rules;
• When F (x) is not that nice, we may have to use definition. These are what you have learned from Differential Calculus (Mat21A or equivalent courses).

Antiderivative problem Question: If the derivative f(x) is given, can you find the original function F(x) such that F ′(x) = f (x)? In other words, If the velocity is given, can you find the position? If the slope of a curve is given, can you figure out the curve?

In general, if F (x) is an antiderivative of f (x) on an interval I, then the most general antiderivative of f on I is

F (x) + C

where C is an arbitrary constant.

• It is a family of functions F (x) + C whose graphs are vertical translations of one another.
• Slopes of all these graphs are equal to f (x).

Indefinite Integral

This family of antiderivatives F (x) + C is very important, we denote it by a special symbol ∫ f (x) dx,

called the indefinite integral of f with respect to x. Here,

• f (x) is the integral;
• x is the variable of integration

is an integral sign. That is, ∫ f (x) dx = F (x) + C, all antiderivatives of f (x).

Example. Find the indefinite integral ∫ x^2 dx.

This is the same as “find antiderivatives of x^2 ”. The answer is ∫ x^2 dx =

x^3 3

+ C.

Some formulas of antiderivatives

The formulas in the table are easily proved by differentiating the antideriv- atives and verifying that the result agrees with the original function.

Indefinite Integral Rules That is, (^) ∫

kf (x) dx = k

f (x) dx ∫ −f (x) dx = −

f (x) dx ∫ f (x) ± g (x) dx =

f (x) dx ±

g (x) dx

e.g. ∫ 4 x^2 dx = 4

x^2 dx = 4 ×

x^3 3

+ C

x^2 + sin xdx =

x^2 dx +

sin xdx =

x^3 3

− cos x + C

Example. Find the indefinite integral ∫ ( 5 x^100 +

x

1 + 4x^2

− 3 e^3 x

dx

∫ ( 5 x^100 +

x

1 + 4x^2

• e^3 x

dx

5 x^100 dx +

x

dx +

1 + 4x^2

dx −

3 e^3 xdx

5 x^101 101

• ln |x| + arctan 2x − e^3 x^ + C.

Example: Solve dy dx

= 3x^2 + sin x y (0) = 1 Solution: y (x) = x^3 − cos x + C Substitute y (0) = 1 yields 1 = y (0) = 0 − cos 0 + C 1 = −1 + C C = 2

Therefore, the solution is

y (x) = x^3 − cos x + 2.

Example: Solve d^2 y d^2 x

= x y (0) = 2, y′^ (0) = 1 Solution: d^2 y d^2 x

= x

yields dy dx

x^2 2

+ C 1.

Using y′^ (0) = 1, we have

02 2

+ C 1 = 1

C 1 = 1.

Example: Find the curve whose slope at the point (x, y) is x^3 + sin (πx) if

the curve is required to pass through the point

1 , 94 + (^) π^1

We need to solve dy dx

= x^3 + sin (πx)

y (1) =

π Answer: y (x) =

x^4 4

cos (πx) π