Applications of Integration

$e^2 - e$

$\ln(e) - \ln(1)$

1

$e - 1$

The area cannot be determined with given information.

Greater than 1 square unit but finite.

Between 0 and 1 square units.

0

$\int_{-3}^{-5}(\lvert((-(\sqrt{x}))+6)-(\sqrt{x}+6)\rvert) dx$

$\int_{3}^{5} ((\sqrt{x}+6)-(-(\sqrt{x}))+6) dx$

$\int_{0}^{9} ((\sqrt{y}+6)-(-(\sqrt{y}))+6) dy$

$\int_{0}^{9}(\lvert g(y)-f(y)\rvert) dy$

$\frac{3}{2}$

$2$

$\frac{\pi}{2}$

$1$

$\int_A(B) [c - d] , dx$

$\int_c^d [B(y) - A(y)] , dy$

$\int_c(d) [A(x) + B(x)] , dx$

$\int_c^d [A(y) - B(y)] , dy$

1/6

1/4

7/60

1/8

$\frac{1}{8}$

$\frac{1}{30}$

$\frac{1}{12}$

$\frac{1}{16}$

$\int_a(b),,,,dxdy,gfdagab$-,,\\[object Object]%^&*()

$\int_a^b [g(y) - f(y)], dy$

$\int_a(b)[f(x)-g(x)]dx, dx$

When $g(y)$ equals zero across all intervals within $[c, d]$.

If $f(y)$ has greater absolute maxima than $g(y)$.

If $f(y)$ is below $g(y)$ on most intervals within $[c, d]$.

When integrating over half-closed intervals such as $[c, d)$.

Integrate $\int_{0}^{1} (y^2 - \sqrt{x}) dx$, although this is incorrect since variables are not consistent and it neglects that we're integrating with respect to $y$.

Integrate $\int_{0}^{1} (\sqrt{y} - y) dx$ since it might seem that we need to solve for $x$, but this does not correctly represent the area between the curves.

Integrate $\int_{0}^{1} (x - y) dy$ as it appears to cover both functions, but incorrectly subtracts them in terms of their dependent variable, not their output.

Integrate $\int_{0}^{1} (y - y^2) dy$ because integrating with respect to $y$ directly incorporates the given range for $y$.