Question

Match each curve to the area under it on the interval [-1, 5]. y = x2 + 16 42 square units y = -x2 + 7x 72 square units y = 4x + 26 204 square units y = -0.5x + 13 138 square units

Answer 1

Answer:

1. $\boxed{y=x^2+16\to138sq.\:units}$

2. $\boxed{y=-x^2+7x\to42sq. \:units}$

3. $\boxed{y=4x+26\to 204sq.\:units}$

4.$\boxed{y=-0.5x+13\to72sq.\:units}$

Step-by-step explanation:

1. The first curve is $y=x^2+16$

The area under this curve on the interval [-1, 5] is given by:

$\int\limits^5_{-1} {x^2+16} \, dx$

We integrate to obtain:

$\frac{1}{3}x^3+16x|_{-1}^5$

We evaluate to obtain:

$\frac{1}{3}(5)^3+16(5)-(\frac{1}{3}(-1)^3+16(-1))=138sq.\:units$

$\boxed{y=x^2+16\to138sq.\:units}$

2. The second curve is $y=-x^2+7x$.

The area under this curve on the interval [-1, 5] is given by:

$\int\limits^5_{-1} {-x^2+7x} \, dx$

We integrate this function to obtain:

$-\frac{1}{3}x^3+\frac{7}{2}x^2|_{-1}^5$

This evaluates to

$-\frac{1}{3}(5)^3+\frac{7}{2}(5)^2-(-\frac{1}{3}(-1)^3+\frac{7}{2}(-1)^2)=42$ square units.

$\boxed{y=-x^2+7x\to42sq. \:units}$

3.  The third curve is $y=4x+26$

The area under this curve on the interval [-1, 5] is given by:

$\int\limits^5_{-1} {4x+26} \, dx$

We integrate this function to obtain:

$2x^2+26x|_{-1}^5$

We evaluate the limits of integration to obtain:

$2(5)^2+26(5)-(2(5)^2+26(5))=204sq.\:units$

$\boxed{y=4x+26\to 204sq.\:units}$

4. The fourth curve is $y=-0.5x+13$

The area under this curve on the interval [-1, 5] is given by:

$\int\limits^5_{-1} {-0,5x+13} \, dx$

We integrate this function to obtain:

$-0.25x^2+13x|_{-1}^5$

We evaluate the limits of integration to obtain:

$-0.25(5)^2+13(5)-(-0.25(-15)^2+13(-1))=72sq.\:units$

$\boxed{y=-0.5x+13\to72sq.\:units}$

Answer 2

Answer: The above answer is correct

Step-by-step explanation: I got this right on Edmentum