Question

Use the Fundamental Theorem of Calculus to find the "area under curve" of y = − x 2 + 7 x between x = 1 and x = 5 . In your calculations, if you need to round, do not do so until the very end of the problem.

Answer 1

Answer:

The area under the curve is of 42.67 area units.

Step-by-step explanation:

Using the Fundamental Theorem of Calculus, the area under a curve f(x), between $x = a$ and $x = b$, is given by:

$A = \int_{a}^{b} f(x) dx$

In this question, we have that:

$f(x) = -x^2 + 7x$, from 1 to 5. So

$A = \int_{a}^{b} f(x) dx$

$A = \int_{1}^{5} (-x^2 + 7x) dx$

$A = -\frac{x^3}{3} + \frac{7x^2}{2}|_1^{5}$

$A = -\frac{5^3}{3} + \frac{7*5^2}{2} + \frac{1^3}{3} - \frac{7*1^2}{2}$

$A = -\frac{124}{3} + \frac{168}{2}$

$A = -\frac{124}{3} + 84$

$A = -\frac{124}{3} + \frac{252}{3}$

$A = \frac{128}{3}$

$A = 42.67$

The area under the curve is of 42.67 area units.

Answer 2

The area under the curve is 42.67 square unit.

Area under curve measures the entire two-dimensional area underneath the entire region of curve.

Area under the curve of f(x), from x = a to x = b is given by,

            $Area=\int\limits^b_a {f(x)} \, dx$

Here given that,   $f(x)=-x^{2} +7x$   and  $a=1,b=5$

Substituting above values in Area integral expression.

                  $Area=\int\limits^5_1 {(-x^{2}+7x) } \, dx \\\\Area = -\frac{x^{3} }{3}+\frac{7}{2}x^{2}$

Substituting the limit of integration.

             $Area=(-\frac{125}{3}+\frac{175}{2} )-(-\frac{1}{3}+\frac{7}{2} )\\\\Area = (-\frac{125}{3}+\frac{1}{3} )+(\frac{175}{2} -\frac{7}{2} )\\\\Area=42.67unit^{2}$

Thus, the area under the curve is 42.67 square unit.

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