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Calculus- Early Transcendentals, 2021a

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242 Integration<br />

Solution. We will use the formula above. We have f (t)=t 3 sin(1 + t), u(x) =10x and v(x) =x 2 .Then<br />

u ′ (x)=10 and v ′ (x)=2x. Thus,<br />

∫<br />

d x 2<br />

t 3 sin(1 +t)dt = (x 2 ) 3 sin(1 +(x 2 ))(2x) − (10x) 3 sin(1 +(10x))(10)<br />

dx 10x<br />

= 2x 7 sin(1 + x 2 ) − 10000x 3 sin(1 + 10x)<br />

♣<br />

Example 6.20: FTC I + Chain Rule<br />

Differentiate the following integral with respect to x:<br />

∫ 2x<br />

x 3<br />

1 + costdt<br />

Solution. Using the formula we have:<br />

∫<br />

d 2x<br />

1 + costdt=(1 + cos(2x))(2) − (1 + cos(x 3 ))(3x 2 ).<br />

dx x 3<br />

♣<br />

Exercises for Section 6.2<br />

Exercise 6.2.1 Evaluate<br />

Exercise 6.2.2 Evaluate<br />

Exercise 6.2.3 Evaluate<br />

Exercise 6.2.4 Evaluate<br />

Exercise 6.2.5 Evaluate<br />

Exercise 6.2.6 Evaluate<br />

∫ 4<br />

1<br />

∫ π<br />

0<br />

∫ 10<br />

1<br />

∫ 5<br />

0<br />

∫ 3<br />

0<br />

∫ 2<br />

1<br />

t 2 + 3tdt<br />

sintdt<br />

1<br />

x dx<br />

e x dx<br />

x 3 dx<br />

x 5 dx<br />

Exercise 6.2.7 Find the derivative of G(x)=<br />

∫ x<br />

1<br />

t 2 − 3tdt

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