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B) The area between the

x

-axis and the curve

y=16-x^{2}

for $1 \leq x \leq 5...

Feb 1, 2024

B) The area between the

x

-axis and the curve

y=16-x^{2}

for

1 \leq x \leq 5

Generated Graph

Solution by Steps

step 1

To find the area between the

x

-axis and the curve

y=16-x^{2}

from

x=1

x=5

, we need to evaluate the definite integral

\int_{1}^{5} (16 - x^2) dx

step 2

The antiderivative of

16 - x^2

16x - \frac{x^3}{3}

step 3

We evaluate the antiderivative at the upper and lower limits of integration and subtract:

\left[16x - \frac{x^3}{3}\right]_{1}^{5}

step 4

Plugging in the upper limit, we get

16(5) - \frac{5^3}{3} = 80 - \frac{125}{3}

step 5

Plugging in the lower limit, we get

16(1) - \frac{1^3}{3} = 16 - \frac{1}{3}

step 6

Subtracting the lower limit from the upper limit, we get

\left(80 - \frac{125}{3}\right) - \left(16 - \frac{1}{3}\right) = 80 - 16 - \left(\frac{125}{3} - \frac{1}{3}\right) = 64 - \frac{124}{3}

step 7

Simplifying the result, we get

64 - 41\frac{1}{3} = 22\frac{2}{3}

\frac{68}{3}

Answer

\frac{68}{3}

or approximately

22.667

Key Concept

Definite Integral as Area

Explanation

The definite integral of a function between two limits gives the area under the curve of the function and above the

x

-axis, between those two limits.