# Answer to Question #6242 in Algebra for Brandy

Question #6242

Use inscribed rectangles to approximate the area under h(x) = -0.5x^2 + 4 for 0 <= x <= 2 and rectangle width 0.25.

Expert's answer

Use inscribed rectangles to approximate the area under h(x) = -0.5x² + 4 for 0 <= x <= 2 and rectangle width 0.25.

First rectangle:

(x1 is the left base point, x2 is the right base point, S1 is rectangle's area)

x1 = 0; x2 = 0.25;

h(x1) = 4; h(x2) = 3.96875;

S1 = min(h(x1),h(x2))*0.25 = 3.96875*0.25 = 0.9921875;

Second rectangle:

x1 = 0.25; x2 = 0.5;

h(x1) = 3.96875; h(x2) = 3.875;

S2 = 0.25*3.875 = 0.96875;

Third rectangle:

x1 = 0.5; x2 = 0.75;

h(x1) = 3.875; h(x2) = 3.71875;

S3 = 0.25*3.71875 = 0.9296875;

Forth rectangle:

x1 = 0.75; x2 = 1;

h(x1) = 3.71875; h(x2) = 3.5;

S4 = 0.25*3.71875 = 0.9296875;

Fifth rectangle:

x1 = 1; x2 = 1.25;

h(x1) = 3.5; h(x2) = 3.21875;

S5 = 0.25*3.21875 = 0.8046875;

Sixth rectangle:

x1 = 1.25; x2 = 1.5;

h(x1) = 3.21875; h(x2) = 2.875;

S6 = 0.25*2.875 = 0.71875;

Seventh rectangle:

x1 = 1.5; x2 = 1.75;

h(x1) = 2.875; h(x2) = 2.46875;

S7 = 0.25*2.46875 = 0.6171875;

x1 = 1.75; x2 = 2;

h(x1) = 2.46875; h(x2) = 2;

S8 = 0.25*2 = 0.5;

So,vapproximated area is S = S1+S2+S3+S4+S5+S6+S7+S8 = 6.4609375.

First rectangle:

(x1 is the left base point, x2 is the right base point, S1 is rectangle's area)

x1 = 0; x2 = 0.25;

h(x1) = 4; h(x2) = 3.96875;

S1 = min(h(x1),h(x2))*0.25 = 3.96875*0.25 = 0.9921875;

Second rectangle:

x1 = 0.25; x2 = 0.5;

h(x1) = 3.96875; h(x2) = 3.875;

S2 = 0.25*3.875 = 0.96875;

Third rectangle:

x1 = 0.5; x2 = 0.75;

h(x1) = 3.875; h(x2) = 3.71875;

S3 = 0.25*3.71875 = 0.9296875;

Forth rectangle:

x1 = 0.75; x2 = 1;

h(x1) = 3.71875; h(x2) = 3.5;

S4 = 0.25*3.71875 = 0.9296875;

Fifth rectangle:

x1 = 1; x2 = 1.25;

h(x1) = 3.5; h(x2) = 3.21875;

S5 = 0.25*3.21875 = 0.8046875;

Sixth rectangle:

x1 = 1.25; x2 = 1.5;

h(x1) = 3.21875; h(x2) = 2.875;

S6 = 0.25*2.875 = 0.71875;

Seventh rectangle:

x1 = 1.5; x2 = 1.75;

h(x1) = 2.875; h(x2) = 2.46875;

S7 = 0.25*2.46875 = 0.6171875;

x1 = 1.75; x2 = 2;

h(x1) = 2.46875; h(x2) = 2;

S8 = 0.25*2 = 0.5;

So,vapproximated area is S = S1+S2+S3+S4+S5+S6+S7+S8 = 6.4609375.

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