Place the following integrals in order of size, starting with the smallest. P = ∫01 2√x dx Q = ∫01 2x dx R = ∫01 (√2)x dx
- AP < Q < R
- BP < R < Q
- CQ < P < R
- DQ < R < P
- ER < P < Q
- FR < Q < P
Show the answer and worked solution
answer · F
- AP < Q < R
- BP < R < Q
- CQ < P < R
- DQ < R < P
- ER < P < Q
- FR < Q < P
Compare the exponents on (0, 1), where 2t is increasing in t. There √x > x > x2, and (√2)x = 2x/2. So 2√ x > 2x > 2x/2 throughout the interval, and integrating preserves the order: R < Q < P.