how do u deal with the 1 in integral(from 0 to 3) (1+sqrt(9-x^2)dx

Question

how to solve with extra addend 1 in the integral?

Answer 1

Hi Rhonda!

To deal with the 1 in this integral, you can break up the original integral as follows:

∫[1+√(9-x^2)]dx = ∫1dx + ∫√(9-x^2)dx

*Note: I have indicated these as indefinite integrals because there is no way to nicely format definite integrals on here.

From there, you take the definite integrals each separately as you normally would, and add them together to find the answer.

Hope this helps! Let me know if you have any further questions.

-Leigh

Answer 2

Rhonda,

This problem can be done much simpler if you notice that the second part of the integral is in the first quadrant and is equal to a quarter of circle's area with r = 3. So,

∫(from 0 to 3) (1+sqrt(9-x^2)dx = 3 + (1/4)pi(3^2)

Answer 3

Remember the rule ∫ ( f(x) + g(x) )dx = ∫ f(x)dx + ∫ g(x)dx. In other words, integration is additive (distributes over addition). That is how the 1 is pulled out (see Leigh's answer.)

If you really wanted to use symbolic integration and plug in the bounds then use trigonometric substitution. Some of you tried to show this but had formatting trouble.

If x = 3 sin θ then dx = 3 cos θ dθ

We get

   ∫ [1+√(9-x²)] dx

= ∫ [1+√(9-9 sin² θ)] * 3 cos θ dθ

= ∫ [3 cos θ + 9 cos² θ] dθ

= ∫ [3 cos θ + 9(1+cos 2θ)/2 ] dθ

= ∫ [3 cos θ + 9/2 + (9/2) cos 2θ] dθ

= 3 sin θ + 9θ/2 + (9/4) sin 2θ + C

= 3 sin θ + 9θ/2 + (9/2) sin θ cos θ+ C

= x + (9/2) sin^-1 (x/3) + (9/2) (x/3) √(1 - x²/9) + C

= x + (9/2) sin^-1 (x/3) + (x/2)√(9 - x²) + C

Substituting the bounds -3 and 3 gives

[3+(9/2) sin^-1 (3/3)+(1/2) (3) √(9 - 3²)] - [0+(9/2) sin^-1 (0/3)+(1/2) (0) √(9 - 0²)]

= [3+(9/2) (π/2) + 0] - [0 + (9/2) (0) + 0]

= 3+9π/4

Same as the answer realized with the recognition that the integral equals the sum of a rectangle and quarter-circle.

Answer 4

Test: θ θ θ Θ