Fundamental Theorem of Calculus: Videos & Practice Problems

The fundamental theorem of calculus connects definite integrals and antiderivatives, allowing for efficient evaluation of integrals. Part one states that if a function is continuous on an interval [a, b], the derivative of its antiderivative equals the original function: $F^{\text{'}}=f\left(x\right)$. Part two enables evaluation of definite integrals by calculating the antiderivative at the bounds: $\int f_{\left(}dx=F\left(b\right)-F\left(a\right)$.

Fundamental Theorem of Calculus Part 1

The fundamental theorem of calculus, particularly part one, establishes a crucial link between derivatives and definite integrals. This theorem asserts that if a function \( f \) is continuous on an interval \([a, b]\), then the derivative of its antiderivative \( F \) is equal to the original function \( f \). In mathematical terms, this can be expressed as:

$$ \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) $$

Here, \( F(x) \) is the antiderivative of \( f(x) \), and the integral represents the area under the curve of \( f(t) \) from \( a \) to \( x \). The theorem essentially states that differentiating the integral of a function yields the function itself, demonstrating that differentiation and integration are inverse processes.

To apply this theorem, one must recognize that when taking the derivative of an integral, the variable of integration (often \( t \)) is replaced with the upper limit of integration (in this case, \( x \)). For example, if we have:

$$ y = \int_{a}^{x} f(t) \, dt $$

then the derivative \( \frac{dy}{dx} \) can be found simply by substituting \( t \) with \( x \), resulting in:

$$ \frac{dy}{dx} = f(x) $$

However, if the upper limit of the integral is a function of \( x \), such as \( g(x) \), the chain rule must be applied. The process involves two steps: first, replace \( t \) with \( g(x) \), and then multiply by the derivative of \( g(x) \). This can be expressed as:

$$ \frac{dy}{dx} = f(g(x)) \cdot g'(x) $$

For instance, if we have:

$$ y = \int_{5}^{x^2} f(t) \, dt $$

the derivative would be calculated as follows:

1. Replace \( t \) with \( x^2 \) to get \( f(x^2) \).

2. Multiply by the derivative of \( x^2 \), which is \( 2x \).

Thus, the final result would be:

$$ \frac{dy}{dx} = f(x^2) \cdot 2x $$

Understanding the fundamental theorem of calculus part one is essential for efficiently solving problems involving derivatives of integrals. It simplifies the process significantly, allowing for quick calculations without the need for lengthy integration followed by differentiation. Mastery of this concept will enhance your ability to tackle a variety of calculus problems effectively.

Given the definite integral $F\left(x\right)={\int }_3^x[t^8-\sin \left(t^4\right)]dt$ , find the derivative $F^{\prime }\left(x\right)$.

• A. $F^{\prime }\left(x\right)=3x^9-3x\sin \left(x^4\right)$
• B. $F^{\prime }\left(x\right)=x^9-x\sin \left(x^4\right)$
• C. $F^{\prime }\left(x\right)=3x^8-3\sin \left(x^4\right)$
• D. $F^{\prime }\left(x\right)=x^8-\sin \left(x^4\right)$

Answer: (D)

Given the definite integral $F\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dh$, find the derivative $F^{\prime}\left(x\right)$.

• A. $F^{\prime }\left(x\right)={20}^5{x}^4+\frac{25,200x}{\sqrt{{\left(20x\right)}^5}}$
• B. $F^{\prime }{\left(x\right)=({20}^{}x)}^4+\frac{1260x}{\sqrt{{\left(20x\right)}^5}}$
• C. $F^{\prime }\left(x\right)={\left(20x\right)}^5+\frac{25,200x^2}{\sqrt{{\left(20x\right)}^5}}$
• D. $F^{\prime }\left(x\right)={\left(20x\right)}^5+\frac{1260x^2}{\sqrt{{\left(20x\right)}^5}}$

Answer: (A)

Fundamental Theorem of Calculus Part 1 Example1

To find the derivative of a definite integral, we can utilize the Fundamental Theorem of Calculus, which simplifies the process significantly. When the upper limit of the integral is a variable, the theorem states that the derivative of the integral with respect to that variable can be computed by substituting the variable into the integrand.

For example, consider the integral:

$$ F(x) = \int_{a}^{x} f(t) \, dt $$

According to the theorem, the derivative is given by:

$$ F'(x) = f(x) $$

This means that if we have an integral with the upper limit as \( x \), we simply replace the variable \( t \) in the integrand with \( x \). For instance, if we have:

$$ F(x) = \int_{a}^{x} (t^4 - 30t^3 + \frac{1}{3}t^2 + 30t - 1000) \, dt $$

Then the derivative \( F'(x) \) would be:

$$ F'(x) = x^4 - 30x^3 + \frac{1}{3}x^2 + 30x - 1000 $$

In cases where the upper limit is a function of \( x \), such as \( g(x) \), we still apply the Fundamental Theorem but must also use the chain rule. For an integral of the form:

$$ F(x) = \int_{a}^{g(x)} f(h) \, dh $$

The derivative is calculated as follows:

$$ F'(x) = f(g(x)) \cdot g'(x) $$

Here, \( f(g(x)) \) is the integrand evaluated at the upper limit, and \( g'(x) \) is the derivative of the upper limit function. For example, if we have:

$$ F(x) = \int_{a}^{x^2 - 13x} (h^2 - 13h^8) \, dh $$

We first evaluate the integrand at \( h = g(x) = x^2 - 13x \):

$$ f(g(x)) = (x^2 - 13x)^2 - 13(x^2 - 13x)^8 $$

Next, we compute the derivative of the upper limit:

$$ g'(x) = 2x - 13 $$

Thus, the final expression for the derivative becomes:

$$ F'(x) = \left((x^2 - 13x)^2 - 13(x^2 - 13x)^8\right)(2x - 13) $$

This approach allows for efficient computation of derivatives of definite integrals, leveraging both the Fundamental Theorem of Calculus and the chain rule when necessary.

Fundamental Theorem of Calculus Part 2

The Fundamental Theorem of Calculus Part Two provides a powerful method for evaluating definite integrals by utilizing antiderivatives. This theorem states that if a function \( f(x) \) is continuous on the interval \([a, b]\), and \( F(x) \) is any antiderivative of \( f(x) \) on that interval, then the definite integral of \( f(x) \) from \( a \) to \( b \) can be computed as:

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]

In this equation, \( F(b) \) represents the value of the antiderivative evaluated at the upper limit \( b \), while \( F(a) \) is the value evaluated at the lower limit \( a \). The process involves finding the antiderivative of the function, substituting the bounds, and then subtracting the results.

For example, to evaluate the integral of \( 2x \) from \( 2 \) to \( 5 \), we first find the antiderivative, which is \( F(x) = x^2 \). We then compute:

\[F(5) - F(2) = 5^2 - 2^2 = 25 - 4 = 21\]

This result indicates that the area under the curve of \( f(x) = 2x \) from \( x = 2 \) to \( x = 5 \) is \( 21 \).

In a more complex example, consider the integral of \( x^2 - 4x + 5 \) from \( 1 \) to \( 4 \). The antiderivative is computed term by term, yielding:

\[F(x) = \frac{x^3}{3} - 2x^2 + 5x\]

Evaluating this from \( 1 \) to \( 4 \) gives:

\[F(4) - F(1) = \left(\frac{4^3}{3} - 2(4^2) + 5(4)\right) - \left(\frac{1^3}{3} - 2(1^2) + 5(1)\right)\]

Calculating each term results in:

\[F(4) = \frac{64}{3} - 32 + 20 = \frac{64}{3} - 12 = \frac{64 - 36}{3} = \frac{28}{3}\]

And for \( F(1) \):

\[F(1) = \frac{1}{3} - 2 + 5 = \frac{1}{3} + 3 = \frac{1 + 9}{3} = \frac{10}{3}\]

Thus, the final evaluation yields:

\[F(4) - F(1) = \frac{28}{3} - \frac{10}{3} = \frac{18}{3} = 6\]

This indicates that the area under the curve of \( f(x) = x^2 - 4x + 5 \) from \( x = 1 \) to \( x = 4 \) is \( 6 \).

In summary, the Fundamental Theorem of Calculus Part Two simplifies the process of finding the area under a curve by connecting definite integrals with antiderivatives, allowing for straightforward calculations when the function is continuous over the specified interval.

Evaluate the following integral:
${\int }_0^{3}4dx$

• A. 12
• B. 4
• C. 3
• D. 0

Answer: (A)

Evaluate the following integral:
${\int }_1^4(2x+1)dx$

• A. 16
• B. 24
• C. 12
• D. 18

Answer: (D)

Evaluate the following integral:
${\int }_{-1}^2(x^2-3x+2)dx$

• A. -6
• B. 0
• C. $\frac{27}{6}$
• D. $\frac{25}{6}$

Answer: (C)

Evaluate the following integral:
${\int }_0^1(2x^3-x^2+4x)dx$

• A. $\frac{13}{6}$
• B. $\frac{1}{6}$
• C. $\frac{17}{6}$
• D. $-\frac{11}{6}$

Answer: (A)

Evaluate the following integral:
${\int }_2^3{x}^{\frac{5}{2}}dx$

• A. 19.360
• B. 16.594
• C. 10.129
• D. 11.817

Answer: (C)

Evaluate the following integral:
${\int }_1^2\frac{5}{x^2}dx$

• A. $\frac{5}{3}$
• B. $\frac{5}{2}$
• C. $\frac{15}{2}$
• D. $5$

Answer: (B)

Fundamental Theorem of Calculus Part 2 Example 2

To evaluate definite integrals, we can utilize the Fundamental Theorem of Calculus, which allows us to find the area under a curve defined by a function over a specific interval. This process involves finding the antiderivative of the function and then applying the bounds of integration.

For example, consider the integral from 0 to 5 of a polynomial function. We start by applying the sum and difference rule for integration, which allows us to integrate each term separately. Using the power rule, we find the antiderivative of each term:

1. For \(x^3\), the antiderivative is \(\frac{x^4}{4}\).

2. For \(3x^2\), it simplifies to \(x^3\) after canceling the coefficient.

3. For \(-6x\), the antiderivative is \(-3x^2\).

4. The constant \(2\) becomes \(2x\).

Combining these results, we have:

\[ F(x) = \frac{x^4}{4} + x^3 - 3x^2 + 2x \]

Next, we evaluate this antiderivative at the upper bound (5) and lower bound (0), and subtract the results:

\[ F(5) - F(0) = \left(\frac{5^4}{4} + 5^3 - 3 \cdot 5^2 + 2 \cdot 5\right) - 0 \]

Calculating this gives us \(\frac{865}{4}\) as the final result for this integral.

In another example, we evaluate the integral of \(\frac{13}{y^3}\) from -2 to 4. We can simplify the integrand by rewriting it as \(13y^{-3}\) and factoring out the constant 13:

\[ \int_{-2}^{4} 13y^{-3} \, dy = 13 \int_{-2}^{4} y^{-3} \, dy \]

Using the power rule, the antiderivative becomes:

\[ F(y) = -\frac{13}{2y^2} \]

We then evaluate this from -2 to 4:

\[ F(4) - F(-2) = -\frac{13}{2 \cdot 4^2} - \left(-\frac{13}{2 \cdot (-2)^2}\right) \]

After simplifying, we find the result to be \(\frac{39}{32}\).

Lastly, consider the integral of \(\frac{3}{\pi} - \cos(\theta)\) from 0 to \(\frac{\pi}{2}\). The antiderivative is:

\[ F(\theta) = \frac{3}{\pi}\theta - \sin(\theta) \]

Evaluating this from 0 to \(\frac{\pi}{2}\) gives:

\[ F\left(\frac{\pi}{2}\right) - F(0) = \left(\frac{3}{\pi} \cdot \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right)\right) - 0 \]

This simplifies to \(\frac{3}{2} - 1 = \frac{1}{2}\).

In summary, the process of evaluating definite integrals involves finding the antiderivative, applying the bounds, and performing arithmetic to arrive at the final result. Mastery of these steps will facilitate solving various integral problems efficiently.

Here’s what students ask on this topic:

What is the Fundamental Theorem of Calculus and why is it important?

The Fundamental Theorem of Calculus is a key principle in calculus that links the concept of differentiation with integration. It consists of two parts. Part one states that if a function is continuous on an interval [a, b], the derivative of its antiderivative equals the original function: $F\text{'}=f\left(x\right)$. Part two allows for the evaluation of a definite integral by calculating the antiderivative at the bounds: $\int f\left(x\right)dx=F\left(b\right)-F\left(a\right)$. This theorem is crucial because it provides a practical way to compute integrals and understand the relationship between the area under a curve and the antiderivative of a function.

How do you apply the Fundamental Theorem of Calculus to evaluate a definite integral?

To apply the Fundamental Theorem of Calculus to evaluate a definite integral, follow these steps: First, ensure the function is continuous on the interval [a, b]. Then, find the antiderivative of the function, denoted as F(x). Once you have the antiderivative, evaluate it at the upper bound b and the lower bound a. The definite integral is the difference between these two values: $\int f\left(x\right)dx=F\left(b\right)-F\left(a\right)$. This process efficiently calculates the area under the curve of the function from a to b.

What is the difference between the first and second parts of the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus consists of two parts, each serving a distinct purpose. Part one establishes the relationship between differentiation and integration, stating that if a function is continuous on an interval [a, b], the derivative of its antiderivative equals the original function: $F\text{'}=f\left(x\right)$. Part two provides a method to evaluate definite integrals by using the antiderivative at the bounds: $\int f\left(x\right)dx=F\left(b\right)-F\left(a\right)$. Essentially, part one connects derivatives and integrals, while part two provides a practical way to compute the value of definite integrals.

How does the chain rule apply to the Fundamental Theorem of Calculus?

The chain rule is applied in the context of the Fundamental Theorem of Calculus when the upper bound of the integral is a function of x, rather than just x itself. In such cases, after finding the antiderivative and substituting the upper bound, you must multiply by the derivative of the upper bound function. For example, if the integral is from a to g(x), the result is $f\left(g\right(x\left)\right)\cdot g\text{'}\left(x\right)$. This application of the chain rule ensures that the derivative of the composite function is correctly accounted for in the evaluation of the integral.

Why is continuity important in the Fundamental Theorem of Calculus?

Continuity is crucial in the Fundamental Theorem of Calculus because it ensures that the function behaves predictably over the interval [a, b]. For part one, continuity guarantees that the antiderivative exists and that the derivative of the antiderivative equals the original function. For part two, continuity ensures that the antiderivative can be evaluated at the bounds without encountering undefined behavior or discontinuities. Without continuity, the relationship between the integral and the antiderivative may not hold, leading to incorrect evaluations of the integral. Thus, continuity is a foundational requirement for the theorem's application.