Fundamental Theorem of Calculus: Videos & Practice Problems

The fundamental theorem of calculus connects definite integrals and antiderivatives. 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 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 \( f(x) \), meaning that \( F'(x) = f(x) \). The theorem essentially states that differentiating the integral of a function from a constant \( a \) to a variable \( x \) yields the function itself evaluated at \( x \). This relationship simplifies the process of finding derivatives of integrals, as it allows for direct substitution of the variable within the integral.

For example, if we have an integral defined as:

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

To find \( \frac{dy}{dx} \), we can simply replace \( t \) with \( x \) in the function \( f \), leading to:

\[\frac{dy}{dx} = f(x)\]

In cases where the upper limit of the integral is a function of \( x \), such as \( g(x) \), the chain rule must be applied. The process involves substituting the upper limit into the integrand and then multiplying by the derivative of the upper limit function. For instance, if we have:

\[y = \int_{5}^{g(x)} f(t) \, dt\]

To find \( \frac{dy}{dx} \), we first replace \( t \) with \( g(x) \) in \( f(t) \), yielding \( f(g(x)) \). Then, we multiply by the derivative of \( g(x) \), resulting in:

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

This application of the chain rule ensures that we account for the changing upper limit when differentiating the integral. Understanding these principles allows for efficient computation of derivatives involving definite integrals, reinforcing the interconnectedness of calculus concepts.

To solidify your understanding, practice applying the fundamental theorem of calculus with various functions and limits, ensuring you are comfortable with both direct substitutions and the chain rule when necessary.

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. $F^{\prime}\left(x\right)=20^5x^4+\frac{25,200x}{\sqrt{\left(20x\right)^5}}$
• B. $F^{\prime}\left(x)=\left(20^{}x\right.\right)^4+\frac{1260x}{\sqrt{\left(20x\right)^5}}$
• 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 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 \( f \) 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 \( b \), and \( F(a) \) is the value of the antiderivative evaluated at the lower limit \( a \). The process involves finding the antiderivative of the function, substituting the bounds into this antiderivative, 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:

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

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 these values 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 calculation is:

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

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:
$\int_0^3\!4\,dx$

• A. 12
• B. 4
• C. 3
• D. 0

Answer: (A)

Evaluate the following integral:
$\int_1^4\!\left(2x+1\right)\,dx$

• A. 16
• B. 24
• C. 12
• D. 18

Answer: (D)

Evaluate the following integral:
$\int_{-1}^2\!\left(x^2-3x+2\right)\,dx$

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

Answer: (C)

Evaluate the following integral:
$\int_0^1\!\left(2x^3-x^2+4x\right)\,dx$

• A. $\frac{13}{6}$
• B. $\frac16$
• C. $\frac{17}{6}$
• D. $-\frac{11}{6}$

Answer: (A)

Evaluate the following integral:
$\int_2^3\!x^{\frac52}\,dx$

• 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. $\frac53$
• B. $\frac52$
• C. $\frac{15}{2}$
• D. $5$

Answer: (B)

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 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, expressed as F' = f. Part two provides a method to evaluate a definite integral by calculating the antiderivative at the bounds, represented as ∫f(x)dx from a to b = F(b) - F(a). This theorem is crucial because it allows for the efficient evaluation of integrals, transforming complex integration problems into simpler differentiation tasks.

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

To apply the Fundamental Theorem of Calculus for evaluating a definite integral, follow these steps: First, ensure the function is continuous on the interval [a, b]. Then, find the antiderivative F(x) of the function f(x). Next, evaluate the antiderivative at the upper bound b and the lower bound a. Finally, subtract the value of the antiderivative at a from its value at b, i.e., F(b) - F(a). This result gives the value of the definite integral ∫f(x)dx from a to b. This process simplifies the evaluation of integrals by leveraging the relationship between differentiation and integration.

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

The Fundamental Theorem of Calculus is divided into two parts, each serving a distinct purpose. Part one establishes the relationship between differentiation and integration, stating that the derivative of the antiderivative of a function equals the original function, F' = f. This part highlights the inverse nature of differentiation and integration. Part two provides a practical method for evaluating definite integrals. It states that the integral of a function over an interval [a, b] can be found by taking the antiderivative at the upper bound and subtracting the antiderivative at the lower bound, expressed as ∫f(x)dx from a to b = F(b) - F(a). While part one focuses on the theoretical connection, part two offers a computational technique.

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 limit of a definite integral is a function of x, rather than just x itself. In such cases, to find the derivative of the integral, you replace the variable inside the integral with the function of x, and then multiply by the derivative of that function. This is an application of the chain rule, which involves taking the derivative of the outer function and multiplying it by the derivative of the inner function. This approach allows for the correct evaluation of the derivative when the upper bound is not a simple variable.

Why is the constant of integration not needed in definite integrals?

In definite integrals, the constant of integration is not needed because it cancels out during the evaluation process. When calculating a definite integral, you find the antiderivative of the function and evaluate it at the upper and lower bounds, then subtract the results. If a constant of integration were included, it would appear in both evaluations and thus cancel out when subtracting, making it unnecessary. This is why, unlike indefinite integrals, definite integrals do not require the addition of a constant of integration, simplifying the computation.