Question

For the following functions \(f\), find the anti-derivative \(F\) that satisfies the given condition. $$f(x)=x^{5}-2 x^{2}+1 ; F(0)=1$$

Short answer

The anti-derivative of the function \(f(x) = x^5 - 2x^2 + 1\) with the given condition \(F(0) = 1\) is \(F(x) = \frac{x^6}{6} - 2\frac{x^3}{3} + x + 1\).

Step-by-step solution

Find the general anti-derivative of the given function

We have the function \(f(x)=x^{5}-2 x^{2}+1\). To find its anti-derivative, we will apply the power rule for integration: \(\int x^{n} dx= \frac{x^{n+1}}{n+1}+C\), where \(n\) is a constant. We have $$\int f(x) dx = \int (x^{5}-2 x^{2}+1) dx$$ Now we integrate term by term: $$F(x) = \int x^{5} dx - 2\int x^{2} dx + \int 1 dx$$ Applying the power rule, we get $$F(x) = \frac{x^{6}}{6} - 2\frac{x^{3}}{3} + x + C$$ Now we need to find the value of \(C\) using the given condition \(F(0) = 1\).

Use the given condition to find the constant term

The given condition is \(F(0) = 1\). We plug in \(x=0\) to find the value of \(C\): $$1 = \frac{0^{6}}{6} - 2\frac{0^{3}}{3} + 0 + C$$ Solving for \(C\), we get $$C = 1$$

Write the final anti-derivative

Now that we have the value of \(C\), we can write the final anti-derivative which satisfies the given condition: $$F(x) = \frac{x^{6}}{6} - 2\frac{x^{3}}{3} + x + 1$$ So the anti-derivative of \(f(x) = x^{5} - 2x^{2} + 1\) with given condition \(F(0) = 1\) is $$F(x) = \frac{x^{6}}{6} - 2\frac{x^{3}}{3} + x + 1$$

Key concepts

Power Rule for Integration

Understanding the Power Rule for Integration is crucial when finding the anti-derivative, or the original function, given its derivative. It's a fundamental technique used repeatedly in calculus.

When you have an expression of the form \( x^n \), where \( n \) is any real number except \( -1 \) (since that calls for a logarithmic function), you can apply this power rule. The rule states that the integral, or the anti-derivative, of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} \), plus an integration constant, symbolized as \( C \). The addition of \( C \) is critical, as it represents any constant value that the derivative would remove, yet is essential for the integrity of the original function.

To illustrate with a clear example: If you integrate \( f(x) = x^5 \), according to the power rule, you get \( F(x) = \frac{x^6}{6} + C \) because you increase the exponent by 1 and divide by the new exponent. This simple operation is at the heart of undoing differentiation and is indispensable in solving calculus problems.

Definite Integration

While the power rule helps us find the general form of an anti-derivative, definite integration allows us to calculate the exact area under a curve between two specific points. Definite integrals have bounds; they're not just expressions with \( + C \); they have numbers describing the interval of integration, typically written as \( \int_{a}^{b} f(x) dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively.

Calculating a definite integral involves finding the anti-derivative of the function and then evaluating it at the upper and lower bounds. Finally, you subtract the value of the anti-derivative at the lower bound from the value at the upper bound. This process gives you the net area between the function's graph and the x-axis within the specified range. In practical terms, it lets you determine quantities like the distance traveled by an object when its velocity function is known, among other applications.

Integration Constant

The integration constant \( C \) is a fundamental aspect of the indefinite integral. Since derivatives of constants are zero, when we find an anti-derivative, we must consider all possible vertical shifts of our function that the derivative wouldn't capture. This is where \( C \) comes in; it's the ‘plus any number’ part of the anti-derivative.

For instance, if you were given an example problem where \( F(0) = 1 \), this specific value allows you to solve for \( C \) by plugging in the known x-value and the condition given. Once you have \( C \) figured out, as seen in the provided step by step solution, you have the exact function that not only follows the general shape determined by the integral but also passes through the specific point required by your initial conditions. This specificity is vital for real-world applications where we need to model behavior that aligns perfectly with observed or initial data.