A) Calculate the area enclosed by the curve y = x² + 1 and the line y = 7 − x. b) Evaluate the following i. ∫[from 2 to 4] (x - 1/x²) dx ii. ∫[from 1 to 3] 5x√(2x² + 7) dx

Rearrange the equation into a standard quadratic form:

Factor the quadratic equation:

The points of intersection are $x = -3$ and $x = 2$. These will be our limits of integration.

Let's test a value, for example, $x=0$:
For the curve: $y = (0)^2 + 1 = 1$.
For the line: $y = 7 - (0) = 7$.
Since $7 > 1$, the line $y = 7 - x$ is above the curve $y = x^2 + 1$ in the interval $[-3, 2]$.

The area $A$ is given by the integral of the upper function minus the lower function.

Substitute the upper limit ($x=2$):

Substitute the lower limit ($x=-3$):

Subtract the lower limit result from the upper limit result:

Find a common denominator (6):

The area enclosed is $\boxed{\frac{125}{6} \text{ square units}}$.

Substitute the upper limit ($x=4$):

Substitute the lower limit ($x=2$):

Subtract the lower limit result from the upper limit result:

The value of the integral is $\boxed{\frac{23}{4}}$.

Let $u = 2x^2 + 7$.
Find the derivative of $u$ with respect to $x$:

Rearrange to find $dx$ in terms of $du$:

We have $5x \, dx$ in the integral, so we can write $x \, dx = \frac{1}{4} du$.
Therefore, $5x \, dx = \frac{5}{4} du$.

When $x = 1$: $u = 2(1)^2 + 7 = 2 + 7 = 9$.
When $x = 3$: $u = 2(3)^2 + 7 = 2(9) + 7 = 18 + 7 = 25$.

Find the antiderivative of $u^{1/2}$:

Calculate the terms:

Substitute these values back:

The value of the integral is $\boxed{\frac{245}{3}}$.