Evaluate the integral by interpreting it in terms of areas. integral: ∫−30(1+9−x2)dx

Question

Evaluate the integral by interpreting it in terms of areas. integral:  ∫−30(1+9−x2)dx

lenkiklisg7 · Accepted Answer

Step 1  First we break the integral into two parts:  ∫−30(1+9−x2)dx=∫−301dx+∫−309−x2dx  The first term ∫−301dx represents area of rectangle with base b=3 and height h=1  ∫−301dx=b×h=3  ∫−309−x2dx represent area of quarter circle with radius r=3  ∫−309−x2dx=14π×32=9π4  ∫−30(1+9−x2)dx=∫−301dx+∫−309−x2dx  ∫−30(1+9−x2)dx=3+9π4

Lakisha Archer · Accepted Answer

Step 1  The given integral is  I=∫−30(1+9−xx)dx  I=∫−301×x+∫−309−x2dx  I=A1+A2  we can plot area A1 as follows    we can plot area A2 as follows    Hence we can calculate the integral as  I=W×H+πr24  I=3×1+π(3)24  I=3+9π4

nick1337 · Accepted Answer

The expression you are having trouble with seems to be incorrect. You have f(x)=1+9−x2 Since a=−3 and b=0 you have Δx=3n and xk=−3+3kn Applying your definition: f(xk)=1+9−(−3+3kn)2 =1+18kn−9k2n2 =1+3n2kn−k2 From this, we see that f(xk)Δx=3n(1+3n2kn−k2) =3n+9n22kn−k2 Then the final expression should be limn⇒∞[∑k=1n(3n+9n22kn−k2)]

Evaluate the integral by interpreting it in terms of areas.

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