Openstax Calculus Vol 1 Ch4 Section 10 Antiderivatives

EXAMPLE 4.52 Evaluating Indefinite Integrals d. ∫tan𝑥cos𝑥𝑑𝑥

Rewrite the integrand as tan𝑥cos𝑥=(sin𝑥/cos𝑥)cos𝑥=sin𝑥. Therefore, ∫tan𝑥cos𝑥=∫sin𝑥=−cos𝑥+𝐶.

CHECKPOINT 4.52

Solve the initial value problem 𝑑𝑦/𝑑𝑥=3𝑥^−2,𝑦(1)=2.

EXAMPLE 4.52 Evaluating Indefinite Integrals c. ∫4/(1+𝑥^2)𝑑𝑥

Using Properties of Indefinite Integrals, write the integral as 4∫1/(1+𝑥^2)𝑑𝑥. Then, use the fact that tan^−1(𝑥) is an antiderivative of 1/(1+𝑥2) to conclude that ∫4/(1+𝑥^2)𝑑𝑥=4tan^−1(𝑥)+𝐶.

CHECKPOINT 4.50

Verify that ∫𝑥cos𝑥𝑑𝑥=𝑥sin𝑥+cos𝑥+𝐶.

antiderivative

A function F is an antiderivative of f if F'(x)=f(x) for all x in the domain of f.

EXAMPLE 4.50 Finding Antiderivatives a. 𝑓(𝑥)=3𝑥^2

Because 𝑑/𝑑𝑥(𝑥3)=3𝑥^2 then 𝐹(𝑥)=𝑥^3 is an antiderivative of 3𝑥^2. Therefore, every antiderivative of 3𝑥^2 is of the form 𝑥^3+𝐶 for some constant 𝐶, and every function of the form 𝑥^3+𝐶 is an antiderivative of 3𝑥^2.

CHECKPOINT 4.51

Evaluate ∫(4𝑥^3−5𝑥^2+𝑥−7)𝑑𝑥.

CHECKPOINT 4.49

Find all antiderivatives of 𝑓(𝑥)=sin𝑥.

EXAMPLE 4.52 Evaluating Indefinite Integrals b. ∫(𝑥^2+4𝑥^(1/3))/x𝑑𝑥

Rewrite the integrand as (𝑥^2+4𝑥^(1/3))/𝑥=𝑥^2/𝑥+4𝑥^(1/3)/𝑥. Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have ∫(𝑥+(4/𝑥^(2/3)))𝑑𝑥=∫𝑥𝑑𝑥+4∫𝑥^(−2/3)𝑑𝑥 =𝑥^2(1/2)+4(1/((−2/3)+1))𝑥^((−2/3)+1)+𝐶 =(1/2𝑥2+12𝑥1/3+𝐶.

EXAMPLE 4.51 Verifying an Indefinite Integral Each of the following statements is of the form ∫𝑓(𝑥)𝑑𝑥=𝐹(𝑥)+𝐶. Verify that each statement is correct by showing that 𝐹′(𝑥)=𝑓(𝑥). a. ∫(𝑥+𝑒^𝑥)𝑑𝑥=((𝑥^2)/2)+(𝑒^𝑥)+𝐶

Since 𝑑/𝑑𝑥(((𝑥^2)/2)+(𝑒^𝑥)+𝐶)=𝑥+𝑒^𝑥, the statement ∫(𝑥+𝑒^𝑥)𝑑𝑥=((𝑥^2)/2)+(𝑒^𝑥)+𝐶 is correct. Note that we are verifying an indefinite integral for a sum. Furthermore, (𝑥^2)/2 and 𝑒^𝑥 are antiderivatives of 𝑥 and 𝑒^𝑥, respectively, and the sum of the antiderivatives is an antiderivative of the sum.

EXAMPLE 4.50 Finding Antiderivatives d. 𝑓(𝑥)=𝑒^𝑥

Since 𝑑/𝑑𝑥(𝑒^𝑥)=𝑒^𝑥, then 𝐹(𝑥)=𝑒^𝑥 is an antiderivative of 𝑒^𝑥. Therefore, every antiderivative of 𝑒^𝑥 is of the form 𝑒^𝑥+𝐶 for some constant 𝐶 and every function of the form 𝑒^𝑥+𝐶 is an antiderivative of 𝑒^𝑥.

EXAMPLE 4.52 Evaluating Indefinite Integrals a. ∫(5𝑥^3−7𝑥^2+3𝑥+4)𝑑𝑥

Using Properties of Indefinite Integrals, we can integrate each of the four terms in the integrand separately. We obtain ∫(5𝑥^3−7𝑥^2+3𝑥+4)𝑑𝑥=∫5𝑥^3𝑑𝑥−∫7𝑥^2𝑑𝑥+∫3𝑥𝑑𝑥+∫4𝑑𝑥. From the second part of Properties of Indefinite Integrals, each coefficient can be written in front of the integral sign, which gives ∫5𝑥^3𝑑𝑥−∫7𝑥^2𝑑𝑥+∫3𝑥𝑑𝑥+∫4𝑑𝑥=5∫𝑥^3𝑑𝑥−7∫𝑥^2𝑑𝑥+3∫𝑥𝑑𝑥+4∫1𝑑𝑥. Using the power rule for integrals, we conclude that ∫(5𝑥^3−7𝑥^2+3𝑥+4)𝑑𝑥=(5/4)𝑥^4−(7/3)𝑥^3+(3/2)𝑥^2+4𝑥+𝐶.

EXAMPLE 4.51 Verifying an Indefinite Integral

Using the product rule, we see that 𝑑𝑑𝑥(𝑥𝑒𝑥−𝑒𝑥+𝐶)=𝑒𝑥+𝑥𝑒𝑥−𝑒𝑥=𝑥𝑒^𝑥. Therefore, the statement ∫𝑥𝑒^𝑥𝑑𝑥=𝑥𝑒^𝑥−𝑒^𝑥+𝐶 is correct. Note that we are verifying an indefinite integral for a product. The antiderivative 𝑥𝑒^𝑥−𝑒^𝑥 is not a product of the antiderivatives. Furthermore, the product of antiderivatives, ((𝑥^2)*(𝑒^𝑥))/2 is not an antiderivative of 𝑥𝑒^𝑥 since 𝑑/𝑑𝑥(((𝑥^2)*(𝑒^𝑥))/2)=𝑥𝑒^𝑥+((𝑥^2)*(𝑒^𝑥))/2≠𝑥𝑒^𝑥. In general, the product of antiderivatives is not an antiderivative of a product.

EXAMPLE 4.50 Finding Antiderivatives c. 𝑓(𝑥)=cos𝑥

We have 𝑑/𝑑𝑥(sin𝑥)=cos𝑥, so 𝐹(𝑥)=sin𝑥 is an antiderivative of cos𝑥. Therefore, every antiderivative of cos𝑥 is of the form sin𝑥+𝐶 for some constant 𝐶 and every function of the form sin𝑥+𝐶 is an antiderivative of cos𝑥.

Decelerating Car A car is traveling at the rate of 88 ft/sec (60 mph) when the brakes are applied. The car begins decelerating at a constant rate of 15 ft/sec^2. a. How many seconds elapse before the car stops?

First we introduce variables for this problem. Let 𝑡 be the time (in seconds) after the brakes are first applied. Let 𝑎(𝑡) be the acceleration of the car (in feet per seconds squared) at time 𝑡. Let 𝑣(𝑡) be the velocity of the car (in feet per second) at time 𝑡. Let 𝑠(𝑡) be the car's position (in feet) beyond the point where the brakes are applied at time 𝑡.�.The car is traveling at a rate of 88ft/sec.88ft/sec. Therefore, the initial velocity is 𝑣(0)=88�(0)=88 ft/sec. Since the car is decelerating, the acceleration is𝑎(𝑡)=−15ft/s2.�(�)=−15ft/s2.The acceleration is the derivative of the velocity,𝑣′(𝑡)=−15.�′(�)=−15.Therefore, we have an initial-value problem to solve:𝑣′(𝑡)=−15,𝑣(0)=88.�′(�)=−15,�(0)=88.Integrating, we find that𝑣(𝑡)=−15𝑡+𝐶.�(�)=−15�+�.Since 𝑣(0)=88,𝐶=88.�(0)=88,�=88. Thus, the velocity function is𝑣(𝑡)=−15𝑡+88.�(�)=−15�+88.To find how long it takes for the car to stop, we need to find the time 𝑡� such that the velocity is zero. Solving −15𝑡+88=0,−15�+88=0, we obtain 𝑡=8815�=8815 sec.

EXAMPLE 4.53 Solving an Initial-Value Problem Solve the initial-value problem 𝑑𝑦/𝑑𝑥=sin𝑥,𝑦(0)=5.

First we need to solve the differential equation. If 𝑑𝑦/𝑑𝑥=sin𝑥, then 𝑦=∫sin(𝑥)𝑑𝑥=−cos𝑥+𝐶. Next we need to look for a solution 𝑦 that satisfies the initial condition. The initial condition 𝑦(0)=5 means we need a constant 𝐶 such that −cos𝑥+𝐶=5. Therefore, 𝐶=5+cos(0)=6. The solution of the initial-value problem is 𝑦=−cos𝑥+6.

THEOREM 4.15 Power Rule for Integrals

For 𝑛≠−1, ∫𝑥^𝑛𝑑𝑥=(𝑥^(𝑛+1)/𝑛+1)+𝐶.

indefinite integral

Given a function 𝑓, the indefinite integral of 𝑓, denoted ∫𝑓(𝑥)𝑑𝑥 is the most general antiderivative of 𝑓. If 𝐹 is an antiderivative of 𝑓, then ∫𝑓(𝑥)𝑑𝑥=𝐹(𝑥)+𝐶. The expression 𝑓(𝑥) is called the integrand and the variable 𝑥 is the variable of integration.

THEOREM 4.16 Properties of Indefinite Integrals

Let 𝐹 and 𝐺 be antiderivatives of 𝑓 and 𝑔, respectively, and let 𝑘 be any real number. Sums and Differences ∫(𝑓(𝑥)±𝑔(𝑥))𝑑𝑥=𝐹(𝑥)±𝐺(𝑥)+𝐶 Constant Multiples ∫𝑘𝑓(𝑥)𝑑𝑥=𝑘𝐹(𝑥)+𝐶

THEOREM 4.14 General Form of an Antiderivative

Let 𝐹 be an antiderivative of 𝑓 over an interval 𝐼. Then, 1. for each constant 𝐶, the function 𝐹(𝑥)+𝐶 is also an antiderivative of 𝑓 over 𝐼; 2. if 𝐺 is an antiderivative of 𝑓 over 𝐼, there is a constant 𝐶 for which 𝐺(𝑥)=𝐹(𝑥)+𝐶 over 𝐼. In other words, the most general form of the antiderivative of 𝑓 over 𝐼 is 𝐹(𝑥)+𝐶.

EXAMPLE 4.50 Finding Antiderivatives b. 𝑓(𝑥)=1/𝑥

Let 𝑓(𝑥)=ln|𝑥|. For 𝑥>0,𝑓(𝑥)=ln(𝑥) and 𝑑/𝑑𝑥(ln𝑥)=1/𝑥. For 𝑥<0,𝑓(𝑥)=ln(−𝑥) and 𝑑/𝑑𝑥(ln(−𝑥))=−(1/−𝑥)=1/𝑥. Therefore, 𝑑/𝑑𝑥(ln|𝑥|)=1/𝑥. Thus, 𝐹(𝑥)=ln|𝑥| is an antiderivative of 1/𝑥. Therefore, every antiderivative of 1/𝑥 is of the form ln|𝑥|+𝐶 for some constant 𝐶 and every function of the form ln|𝑥|+𝐶 is an antiderivative of 1/𝑥.