#### To determine

**To evaluate:**

∫15dtt-42 if it exists

#### Answer

Does not exist

#### Explanation

**1) Concept:**

If the function has a infinite discontinuity on the given interval [a, b] then the integral does not exist over the interval.

**2) Given:**

∫15dtt-42

**3) Calculation:**

Consider,

∫15dtt-42

Consider the denominator t-42 and equate with 0

t-42=0 becomes t=4

The function has infinite discontinuity at t=4

Therefore, the given function has infinite discontinuity on the interval [1, 5]

Hence, the integration does not exist

**Conclusion:**

Therefore,

∫15dtt-42 does not exist