Antiderivative, Integrals, And Area Calculation: Algebra Problems

Hey guys! Let's dive into some cool algebra problems today. We're going to tackle finding antiderivatives, evaluating integrals, and calculating areas. So, buckle up and let's get started!

Finding the Antiderivative

Let's start with the first part: finding the antiderivative F(x) that satisfies F(1) = 3 for the function y = 3/(x^2) + x^2 - x + 3. This might sound intimidating, but it's totally manageable. The key here is understanding what an antiderivative is and how to find it.

What is an Antiderivative?

In simple terms, an antiderivative is the reverse process of differentiation. If you have a function, say f(x), its antiderivative F(x) is a function whose derivative is f(x). Mathematically, this means F'(x) = f(x). Think of it like going backward in calculus. Differentiation breaks down a function into its rate of change, while antidifferentiation builds a function from its rate of change.

Finding the Antiderivative of y = 3/(x^2) + x^2 - x + 3

Okay, so we need to find F(x) such that F'(x) = 3/(x^2) + x^2 - x + 3. Let’s break this down term by term. Remember, the power rule for integration is ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. We’ll use this a lot here.

1. Term 1: 3/(x^2)

   This can be rewritten as 3x^(-2). Applying the power rule, we get:

   ∫3x^(-2) dx = 3 ∫x^(-2) dx = 3 * (x^(-1)/(-1)) + C = -3/x + C

2. Term 2: x^2

   Applying the power rule again:

   ∫x^2 dx = (x^3)/3 + C

3. Term 3: -x

   This is the same as -x^1. Using the power rule:

   ∫-x dx = - ∫x^1 dx = -(x^2)/2 + C

4. Term 4: 3

   The integral of a constant is simply the constant times x:

   ∫3 dx = 3x + C

Now, let’s put it all together. The antiderivative F(x) is:

F(x) = -3/x + (x^3)/3 - (x^2)/2 + 3x + C

Finding the Constant of Integration

We're not quite done yet! We have a + C hanging around. This is the constant of integration, and it represents the fact that there are infinitely many antiderivatives that differ only by a constant. But, we have a condition: F(1) = 3. This will help us find the specific value of C.

Plug in x = 1 into our antiderivative:

F(1) = -3/1 + (1^3)/3 - (1^2)/2 + 3(1) + C = 3

Simplify:

-3 + 1/3 - 1/2 + 3 + C = 3

1/3 - 1/2 + C = 3

Now, let's solve for C:

C = 3 - 1/3 + 1/2 = 3 - 2/6 + 3/6 = 3 + 1/6 = 19/6

So, the specific antiderivative we’re looking for is:

F(x) = -3/x + (x^3)/3 - (x^2)/2 + 3x + 19/6

Boom! We found the antiderivative that satisfies the given condition. See? Not so scary when we break it down.

Evaluating Integrals

Next up, let's tackle some integrals. Integrals are fundamental to calculus and have tons of applications in physics, engineering, and more. We'll look at two integrals today:

1. ∫(6-7x) dx
2. ∫cos(24x) dx from 0 to π/12

Integral 1: ∫(6-7x) dx

This is a straightforward indefinite integral. Remember, an indefinite integral gives us a general function, whereas a definite integral (like the second one) gives us a numerical value.

To evaluate this integral, we’ll use the power rule and the constant multiple rule (∫cf(x) dx = c∫f(x) dx). Let’s break it down:

∫(6-7x) dx = ∫6 dx - ∫7x dx

Now, integrate each term:

∫6 dx = 6x + C1

∫7x dx = 7 ∫x dx = 7 * (x^2)/2 + C2 = (7x^2)/2 + C2

Combine them:

6x - (7x^2)/2 + C

Where C = C1 - C2 is the constant of integration. So, the result is:

∫(6-7x) dx = 6x - (7x^2)/2 + C

Integral 2: ∫cos(24x) dx from 0 to π/12

This is a definite integral, which means we'll end up with a numerical answer. We're integrating cos(24x) from 0 to π/12. This involves a u-substitution. Let’s do it!

Let u = 24x. Then, du = 24 dx, so dx = du/24.

Now, we rewrite the integral in terms of u:

∫cos(24x) dx = ∫cos(u) (du/24) = (1/24) ∫cos(u) du

The integral of cos(u) is sin(u), so:

(1/24) ∫cos(u) du = (1/24) sin(u) + C

Now, substitute back u = 24x:

(1/24) sin(24x) + C

We have the antiderivative, but we need to evaluate the definite integral from 0 to π/12. So, we plug in the limits of integration:

[(1/24) sin(24 * π/12)] - [(1/24) sin(24 * 0)]

Simplify:

(1/24) sin(2π) - (1/24) sin(0)

Since sin(2π) = 0 and sin(0) = 0, the result is:

(1/24) * 0 - (1/24) * 0 = 0

So, the value of the definite integral is:

∫cos(24x) dx from 0 to π/12 = 0

Finding the Area of a Plane Figure

Last but not least, let’s find the area of the plane figure bounded by the lines y = 3/(x^2), x = 1, x = 3, and y = -4. This is a classic application of integrals.

Visualizing the Problem

First, it's always helpful to visualize what's going on. We have a curve y = 3/(x^2), two vertical lines x = 1 and x = 3, and a horizontal line y = -4. The area we’re interested in is the region enclosed by these lines.

Setting up the Integral

The area between two curves, f(x) and g(x), from x = a to x = b is given by:

Area = ∫[a to b] |f(x) - g(x)| dx

In our case, f(x) = 3/(x^2) and g(x) = -4. The limits of integration are x = 1 and x = 3. So, the integral becomes:

Area = ∫[1 to 3] |3/(x^2) - (-4)| dx = ∫[1 to 3] (3/(x^2) + 4) dx

Evaluating the Integral

Now, we need to evaluate this definite integral. First, let’s find the antiderivative of 3/(x^2) + 4:

∫(3/(x^2) + 4) dx = ∫(3x^(-2) + 4) dx

Integrate term by term:

∫3x^(-2) dx = 3 ∫x^(-2) dx = 3 * (x^(-1)/(-1)) = -3/x

∫4 dx = 4x

So, the antiderivative is:

-3/x + 4x + C

Now, we evaluate the definite integral from 1 to 3:

[(-3/3 + 43)] - [(-3/1 + 41)]

Simplify:

(-1 + 12) - (-3 + 4)

11 - 1 = 10

So, the area of the plane figure is:

Area = 10 square units

Conclusion

Alright, guys, we've covered a lot today! We tackled finding antiderivatives, evaluating integrals (both indefinite and definite), and calculated the area of a plane figure. These are fundamental concepts in calculus, and mastering them will set you up for success in more advanced topics. Keep practicing, and you'll become calculus pros in no time! Remember, every problem is just a puzzle waiting to be solved. Keep that mindset, and you'll nail it!