Calculus- Early Transcendentals, 2021a

6.2. The Fundamental Theorem of Calculus 241
Example 6.18: Using FTC
Differentiate the following function:
g(x)=
∫ x
−2
cos(1 + 5t)sintdt.
Solution. We simply apply the Fundamental Theorem of Calculus directly to get:
g ′ (x)=cos(1 + 5x)sinx.
♣
Using the Chain Rule we can derive a formula for some more complicated problems. We have:
∫
d v(x)
f (t)dt = f (v(x)) · v ′ (x).
dx a
Now what if the upper limit is constant and the lower limit is a function of x? Then we interchange the
limits and add a minus sign to get:
∫
d a
f (t)dt = − d ∫ u(x)
f (t)dt = − f (u(x)) · u ′ (x).
dx u(x) dx a
Combining these two we can get a formula where both limits are a function of x. We break up the
integral as follows:
∫ v(x) ∫ a ∫ v(x)
f (t)dt = f (t)dt + f (t)dt.
u(x)
u(x)
a
We just need to make sure f (a) exists after we break up the integral. Then differentiating and using the
above two formulas gives:
FTC I + Chain Rule
∫
d v(x)
f (t)dt = f (v(x))v ′ (x) − f (u(x))u ′ (x)
dx u(x)
Many textbooks do not show this formula and instead to solve these types of problems will use FTC I
along with the tricks we used to derive the formula above. Either method is perfectly fine.
Example 6.19: FTC I + Chain Rule
Differentiate the following integral:
∫ x 2
10x
t 3 sin(1 +t)dt.