Integration and Applications of Integration in Calculus

Integration and Its Applications

Finding a Function from Its Derivative (Initial Value Problems)

Given the derivative of a function, we can recover the original function by integrating and applying an initial value to determine the constant of integration.

• Key Point: If \( f'(x) = g(x) \) and \( f(a) = b \), then \( f(x) = \int g(x) \, dx + C \), and use the initial value to solve for \( C \).

• Example: If \( f'(x) = 2x \) and \( f(0) = 3 \), then \( f(x) = x^2 + C \). Using the initial value: \( 3 = 0^2 + C \Rightarrow C = 3 \), so \( f(x) = x^2 + 3 \).

Approximate Displacement Using Subintervals

Displacement can be approximated by dividing the interval into subintervals and summing the products of velocity and time width for each subinterval (Riemann sums).

• Key Point: For velocity function \( v(t) \) over \( [a, b] \), approximate displacement using \( n \) subintervals: $\sum_{i=1}^n v(t_i^*) \Delta t$

• Example: If \( v(t) = t \) on \( [0, 2] \) with 2 subintervals, \( \Delta t = 1 \), and sample points \( t_1^* = 0.5, t_2^* = 1.5 \): $v(0.5) \cdot 1 + v(1.5) \cdot 1 = 0.5 + 1.5 = 2$

Properties of Definite Integrals

Definite integrals have several useful properties that simplify calculations and allow for manipulation of integrals.

• Linearity: $\int_a^b [cf(x) + dg(x)] \, dx = c\int_a^b f(x) \, dx + d\int_a^b g(x) \, dx$

• Additivity: $\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx$

• Reversal of Limits: $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$

Derivatives of Definite Integrals

The derivative of a definite integral with a variable upper limit is given by the Fundamental Theorem of Calculus, Part 1.

• Key Point: If \( F(x) = \int_a^x f(t) \, dt \), then $F'(x) = f(x)$

• Example: $\frac{d}{dx} \int_0^x \sin t \, dt = \sin x$

Evaluating Definite Integrals Using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, allowing us to evaluate definite integrals using antiderivatives.

• Key Point: $\int_a^b f(x) \, dx = F(b) - F(a)$, where \( F \) is any antiderivative of \( f \).

• Example: $\int_1^3 x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_1^3 = \frac{1}{3}(27) - \frac{1}{3}(1) = 9 - \frac{1}{3} = \frac{26}{3}$

Using Symmetry to Evaluate or Simplify Definite Integrals

Symmetry properties of functions can simplify the evaluation of definite integrals, especially for even and odd functions over symmetric intervals.

• Even Function: $\int_{-a}^a f(x) \, dx = 2\int_0^a f(x) \, dx$ if \( f(x) \) is even.

• Odd Function: $\int_{-a}^a f(x) \, dx = 0$ if \( f(x) \) is odd.

Average Value of a Function

The average value of a function over an interval is found by integrating the function and dividing by the interval's length.

• Formula: $f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx$

• Example: For \( f(x) = x^2 \) on \( [0, 2] \): $f_{avg} = \frac{1}{2-0} \int_0^2 x^2 \, dx = \frac{1}{2} \left[ \frac{1}{3}x^3 \right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$

Integration by Substitution

Substitution is a method for evaluating integrals by changing variables to simplify the integrand.

• Indefinite Integrals: Let \( u = g(x) \), then $\int f(g(x))g'(x) \, dx = \int f(u) \, du$

• Definite Integrals: Change the limits of integration to match the new variable.

• Example: $\int_0^1 2x \cos(x^2) \, dx$ Let \( u = x^2 \), \( du = 2x \, dx \), limits: \( x=0 \Rightarrow u=0, x=1 \Rightarrow u=1 \): $\int_0^1 \cos(u) \, du = \sin(u) \Big|_0^1 = \sin(1) - \sin(0) = \sin(1)$

Displacement and Position from Velocity

Displacement is the net change in position, found by integrating velocity. Position is found by adding displacement to the initial position.

• Displacement: $\int_a^b v(t) \, dt$

• Position: $s(b) = s(a) + \int_a^b v(t) \, dt$

Area Between Two Curves

The area between two curves is found by integrating the difference of the functions over the interval where one is above the other.

• Formula: $\int_a^b [f(x) - g(x)] \, dx$, where \( f(x) \geq g(x) \) on \( [a, b] \).

• Example: Area between \( y = x^2 \) and \( y = x \) on \( [0, 1] \): $\int_0^1 (x - x^2) \, dx = \left[ \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Volumes of Solids of Revolution: Disk and Washer Methods

These methods are used to find the volume of a solid formed by revolving a region around an axis.

• Disk Method: $V = \pi \int_a^b [R(x)]^2 \, dx$

• Washer Method: $V = \pi \int_a^b \left( [R_{outer}(x)]^2 - [R_{inner}(x)]^2 \right) \, dx$

• About Non-Axis Lines: Adjust the radius expressions to measure distance from the curve to the line of revolution.

Volumes of Solids of Revolution: Shell Method

The shell method finds volume by integrating cylindrical shells formed by revolving a region about an axis.

• Formula (about y-axis): $V = 2\pi \int_a^b x f(x) \, dx$

• About Non-Axis Lines: Adjust the radius to be the distance from the shell to the line of revolution.

Comparing Disk/Washer and Shell Methods

Both methods compute the same volume but are more convenient in different situations depending on the axis of revolution and the functions involved.

• Disk/Washer: Slices perpendicular to the axis of revolution.

• Shell: Slices parallel to the axis of revolution.

• Example: For \( y = x^2 \) revolved about the y-axis, shell method is often easier.

Finding Volumes Using Both Methods

Some problems can be solved using either the disk/washer or shell method, and comparing results can verify correctness.

• Key Point: Set up both integrals and confirm they yield the same volume.