Warm Up

∫3.2

4.8

sin 𝑥 dx

7.1 A – Initial Value Problems

Consider: 2 3y x

then: 2y x 2y x

2 5y x or

It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears.

However, when we try to reverse the operation:

Given: 2y x find y

2y x C

We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant.

Think about it….

Integrals such as are called definite

integrals because we can find a definite value

for the answer.

4 2

1x dx∫

43

1

1

3x C

3 31 14 1

3 3C C

64 1

3 3C C

63

3 21

The constant always cancels when finding a definite integral, so we leave it out!

So far, we have found definite integrals.

4 2

1x dx∫

Integrals such as are called indefinite integrals

because we can not find a definite value for the answer.

2x dx∫

2x dx∫When finding indefinite integrals, we always include the “plus C”.

Indefinite Integrals

31

3x C¿

Common antiderivatives YOU NEED TO KNOW

∫ dxxn.1 1,1

1

nCn

xn

∫x

dx.2 Cx ln

∫ dxex.3 Cex

∫ dxxsin.4 Cx cos

More...

∫ dxxcos.5

∫ dxx2sec.6

∫ dxx2csc.7

∫ dxxx tansec.8

Cx sin

Cx sec

Cx cot

Cx tan

∫ dxxx cotcsc.9 Cx csc

Examples

∫ dxxx )32( 3 Cxxx

34

24

∫ dxxx3Cx 2

9

9

2

∫

dttt

cos12 Ct

t sin

1

More Examples

∫ dxxxx )cos(tansec Cxx sec

∫ dxxx )cscsin1( 2 Cxx cos

∫ dtte t )( 22 Cte t

32

32

Initial Value Problems

We cannot find C unless additional information is given. All we need is a single point, called an initial condition. A problem with an initial condition is called a initial value problem.

Example

• Find the solution to the equation whose graph passes through the point (1,0).

• Apply the initial condition (plug the point in for x and y)• • Now Solve for C

• So our solution is

Example

Find the position function for v(t) = t3 - 2t2 + t s(0) = 1

Cttt

tvts ∫ 23

2

4)()(

234

C2

)0(

3

)0(2

4

01

234C = 1

123

2

4)(

234

ttt

ts

Homework