Engineering Board Exam Reviewer · Integration by Parts · hard

Evaluate: x ≤ft(x ),dx (for 0<x<1)

1. - x 1 - x24 + ≤ft(2 x2 - 1 ) ≤ft(x )4+C
2. x 1 - x24 + ≤ft(2 x2 - 1 ) ≤ft(x )4+C
3. x2 ≤ft(x )2 + x 1 - x24+C
4. x 1 - x24 + ≤ft(2 x2 + 1 ) ≤ft(x )4+C

Answer

B. x 1 - x24 + ≤ft(2 x2 - 1 ) ≤ft(x )4+C

Solution

Let u= x, dv=x,dx: I=x2/2 x- 12 x2sqrt(1-x2),dx. With x= , x2sqrt(1-x2)dx= 12 x- xsqrt(1-x2)2. Hence I=2x2-1/4 x+ xsqrt(1-x2)4+C.

More Integration by Parts questions

1. Evaluate: ≤ft( x ),dx (assume x>0)
2. Evaluate: x3 ex2,dx
3. Evaluate: ≤ft( x )2,dx (assume x>0)
4. Evaluate: ≤ft(x ),dx
5. Evaluate: e2 x ≤ft(3 x ),dx
6. Evaluate: x2 e3 x,dx
7. Evaluate: x2 ≤ft(x ),dx
8. Evaluate: x ≤ft(2 x ),dx
9. Evaluate: x3 x,dx (assume x>0)
10. Evaluate: x,ex,dx

All Integration by Parts questions