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For each problem approximate the area under the curve under the given interval using five trapezoids.

Please help! For each problem approximate the area under the curve under the given-example-1

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Answer:

area ≈ 9.219 square units

Explanation:

You want the approximate area under the curve y = -1/2x² +x +5 on the interval [1.5, 4] using 5 trapezoids.

Trapezoid area

The interval can be divided into 5 intervals of width ...

(4 -1.5)/5 = 2.5/5 = 0.5

The "bases" of each trapezoid will be the function values at the ends of the intervals, for example, at x=1.5 and x=2. The "height" of each trapezoid is the width of the sub-interval, 0.5.

The area formula for a trapezoid applies:

A = 1/2(b1 +b2)h

A = 1/2(f(x) +f(x +0.5))·0.5 . . . . . for x = 1.5, 2, 2.5, 3, 3.5

Approximate total area

The sum of the areas is computed in the attachment as ...

area under the curve = 9.21875

__

Additional comment

The value of the integral is 445/48 ≈ 9.2708333...

Please help! For each problem approximate the area under the curve under the given-example-1
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User Justin K
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