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| Evaluation of Definite Integrals using Fundamental Theorem | Evaluation of Definite Integrals by Substitution | Properties of Definite Integrals and their Application in Evaluation |
Definite Integrals: Evaluation and Properties
Evaluation of Definite Integrals using Fundamental Theorem
The most efficient and widely used method for evaluating definite integrals is provided by the Second Fundamental Theorem of Calculus. This theorem establishes a direct link between the process of finding antiderivatives (indefinite integration) and the numerical value of the definite integral (representing net signed area).
The Evaluation Theorem (Second Fundamental Theorem of Calculus - FTC2)
Statement: If $f$ is a continuous function on a closed interval $[a, b]$, and $F$ is any antiderivative of $f$ on the same interval $[a, b]$ (meaning $F'(x) = f(x)$ for all $x$ in $[a, b]$), then the definite integral of $f$ from $a$ to $b$ is given by the difference of the values of the antiderivative evaluated at the upper and lower limits of integration.
$\int_{a}^{b} f(x) dx = F(b) - F(a)$
Notation: The difference $F(b) - F(a)$ is often expressed using the notation $[F(x)]_{a}^{b}$ or $F(x) \bigg|_{a}^{b}$. The brackets or vertical bar indicate that you should evaluate the expression inside at the upper limit and subtract the value of the expression evaluated at the lower limit.
So, the Second Fundamental Theorem is frequently written as:
$\int_{a}^{b} f(x) dx = [F(x)]_{a}^{b} = F(b) - F(a)$
Significance: This theorem is fundamental because it provides a practical way to compute definite integrals without having to deal with the complex limit of Riemann sums definition. It highlights that finding an antiderivative is key to evaluating definite integrals. The constant of integration $C$ from indefinite integrals is not needed here, as it cancels out in the subtraction: $[F(x)+C]_{a}^{b} = (F(b)+C) - (F(a)+C) = F(b)+C - F(a)-C = F(b)-F(a)$.
Steps for Evaluation using FTC2
To evaluate a definite integral $\int_{a}^{b} f(x) dx$ using the Second Fundamental Theorem of Calculus:
- Verify Continuity: Confirm that the integrand function $f(x)$ is continuous on the closed interval of integration $[a, b]$. If it is not, FTC2 in this form might not apply directly, and the integral might be improper or not exist.
- Find an Antiderivative: Find any one antiderivative $F(x)$ of the integrand $f(x)$. This means finding a function $F(x)$ such that $\frac{d}{dx}F(x) = f(x)$. Use standard integration formulas and techniques (like the power rule, trigonometric integrals, exponential integrals, substitution, integration by parts, etc.) to find the indefinite integral $\int f(x) dx$, and choose a particular antiderivative (e.g., set the constant of integration $C=0$).
- Evaluate the Antiderivative at the Upper Limit: Calculate the value of the antiderivative function $F(x)$ when $x$ is equal to the upper limit of integration, $b$. This value is $F(b)$.
- Evaluate the Antiderivative at the Lower Limit: Calculate the value of the antiderivative function $F(x)$ when $x$ is equal to the lower limit of integration, $a$. This value is $F(a)$.
- Subtract the Values: Compute the difference $F(b) - F(a)$. This numerical difference is the value of the definite integral.
Example 1. Evaluate $\int_{0}^{\pi/2} \cos x dx$.
Answer:
We need to evaluate the definite integral $\int_{0}^{\pi/2} \cos x dx$. The integrand is $f(x) = \cos x$, which is continuous on the interval $[0, \pi/2]$. We can use the Second Fundamental Theorem of Calculus.
Step 1: Verify Continuity.
The function $f(x) = \cos x$ is continuous for all real numbers, so it is continuous on the closed interval $[0, \pi/2]$. FTC2 applies.
Step 2: Find an Antiderivative $F(x)$ of $f(x) = \cos x$.
We need a function $F(x)$ such that $F'(x) = \cos x$. From standard integration formulas, we know that $\int \cos x dx = \sin x + C$. We can choose $F(x) = \sin x$ (by setting $C=0$).
Step 3 & 4: Evaluate $F(x)$ at the Limits.
Evaluate $F(x) = \sin x$ at the upper limit $b=\pi/2$ and the lower limit $a=0$.
"$F(\pi/2) = \sin(\pi/2) = 1$"
[Evaluate at upper limit]
"$F(0) = \sin(0) = 0$"
[Evaluate at lower limit]
Step 5: Subtract the Values.
Compute the difference $F(\pi/2) - F(0)$.
"$\int_{0}^{\pi/2} \cos x dx = F(\pi/2) - F(0) = 1 - 0 = 1$"
Using the bracket notation:
"$\int_{0}^{\pi/2} \cos x dx = [\sin x]_{0}^{\pi/2}$"
[Antiderivative from 0 to $\pi/2$]
"$= \sin(\pi/2) - \sin(0)$"
[Evaluate at upper limit - evaluate at lower limit]
"$= 1 - 0 = 1$"
The value of the definite integral $\int_{0}^{\pi/2} \cos x dx$ is 1.
Example 2. Evaluate $\int_{1}^{e} \frac{1}{x} dx$.
Answer:
We need to evaluate $\int_{1}^{e} \frac{1}{x} dx$. The integrand is $f(x) = \frac{1}{x}$. The interval is $[1, e]$. Note that $e \approx 2.718$, so $1 < e$. The function $f(x) = \frac{1}{x}$ is defined and continuous for all $x \neq 0$. Since the interval $[1, e]$ does not contain 0, $f(x)$ is continuous on this closed interval. FTC2 applies.
Step 1: Verify Continuity.
$f(x) = 1/x$ is continuous on $[1, e]$ since $0 \notin [1, e]$.
Step 2: Find an Antiderivative $F(x)$ of $f(x) = \frac{1}{x}$.
We need $F'(x) = \frac{1}{x}$. From standard integration formulas, $\int \frac{1}{x} dx = \ln |x| + C$. Since the interval of integration $[1, e]$ contains only positive values of $x$, $|x| = x$. We can choose $F(x) = \ln x$ (by setting $C=0$).
Step 3 & 4: Evaluate $F(x)$ at the Limits.
Evaluate $F(x) = \ln x$ at the upper limit $b=e$ and the lower limit $a=1$.
"$F(e) = \ln e = 1$"
[Evaluate at upper limit]
(Recall $\ln e = 1$ because $e^1 = e$).
"$F(1) = \ln 1 = 0$"
[Evaluate at lower limit]
(Recall $\ln 1 = 0$ because $e^0 = 1$).
Step 5: Subtract the Values.
Compute $F(e) - F(1)$.
"$\int_{1}^{e} \frac{1}{x} dx = F(e) - F(1) = 1 - 0 = 1$"
Using the bracket notation:
"$\int_{1}^{e} \frac{1}{x} dx = [\ln x]_{1}^{e}$"
[Antiderivative from 1 to e]
"$= \ln e - \ln 1$"
[Evaluate at upper limit - evaluate at lower limit]
"$= 1 - 0 = 1$"
The value of the definite integral $\int_{1}^{e} \frac{1}{x} dx$ is 1.
Example 3. Evaluate $\int_{0}^{1} (4x^3 + e^x) dx$.
Answer:
We need to evaluate $\int_{0}^{1} (4x^3 + e^x) dx$. The integrand is $f(x) = 4x^3 + e^x$. The interval is $[0, 1]$. The function $f(x)$ is a sum of a polynomial and an exponential function, both of which are continuous everywhere. Thus, $f(x)$ is continuous on $[0, 1]$. FTC2 applies.
Step 1: Verify Continuity.
$f(x) = 4x^3 + e^x$ is continuous on $[0, 1]$.
Step 2: Find an Antiderivative $F(x)$ of $f(x) = 4x^3 + e^x$.
We need $F'(x) = 4x^3 + e^x$. Use linearity (Sum Rule and Constant Multiple Rule) and standard formulas:
"$\int (4x^3 + e^x) dx = \int 4x^3 dx + \int e^x dx$"
$= 4 \int x^3 dx + \int e^x dx$"
Using the Power Rule for $x^3$ ($\int x^3 dx = \frac{x^4}{4} + C_1$) and the standard integral for $e^x$ ($\int e^x dx = e^x + C_2$):
$= 4 \left(\frac{x^4}{4}\right) + e^x + C$"
[Substitute integrals and add C]
"$F(x) = x^4 + e^x$"
[Choosing $C=0$ for $F(x)$]
Step 3 & 4: Evaluate $F(x)$ at the Limits.
Evaluate $F(x) = x^4 + e^x$ at the upper limit $b=1$ and the lower limit $a=0$.
"$F(1) = (1)^4 + e^1 = 1 + e$"
[Evaluate at upper limit]
"$F(0) = (0)^4 + e^0 = 0 + 1 = 1$"
[Evaluate at lower limit]
(Recall $e^0 = 1$).
Step 5: Subtract the Values.
Compute $F(1) - F(0)$.
"$\int_{0}^{1} (4x^3 + e^x) dx = F(1) - F(0) = (1 + e) - 1$"
"$= 1 + e - 1 = e$"
Using the bracket notation:
"$\int_{0}^{1} (4x^3 + e^x) dx = [x^4 + e^x]_{0}^{1}$"
[Antiderivative from 0 to 1]
"$= (1^4 + e^1) - (0^4 + e^0)$"
[Evaluate at upper limit - evaluate at lower limit]
"$= (1 + e) - (0 + 1) = 1 + e - 1 = e$"
The value of the definite integral $\int_{0}^{1} (4x^3 + e^x) dx$ is e.
Evaluation of Definite Integrals by Substitution
When evaluating a definite integral $\int_{a}^{b} f(x) dx$ where the corresponding indefinite integral requires the substitution method (u-substitution), we have two main approaches:
- Method 1: Evaluate the Indefinite Integral First. Find the indefinite integral $\int f(x) dx$ using substitution. Make sure to substitute back to the original variable $x$ to get the antiderivative $F(x)$. Then apply the Second Fundamental Theorem of Calculus using the original limits $a$ and $b$, i.e., calculate $F(b) - F(a)$.
- Method 2: Change the Limits of Integration. Perform the substitution but also transform the limits of integration from the original variable $x$ to the new variable $u$. Then, evaluate the transformed integral with the new limits using FTC2 directly, without needing to substitute back to $x$. This method is often more efficient, especially when the substitution is complex.
Method 2 is generally preferred and is described in detail below.
Method 2: Changing the Limits of Integration During Substitution
Suppose we want to evaluate the definite integral $\int_{a}^{b} f(g(x)) g'(x) dx$, where a suitable substitution is $u = g(x)$.
- Choose the Substitution: Identify the substitution $u = g(x)$ and find the differential $du = g'(x) dx$.
- Change the Limits of Integration: Determine the new limits of integration that correspond to the variable $u$. These limits are obtained by plugging the original limits for $x$ into the substitution equation $u = g(x)$.
- The new lower limit for $u$ is $u_{\text{lower}} = g(a)$ (the value of $u$ when $x=a$).
- The new upper limit for $u$ is $u_{\text{upper}} = g(b)$ (the value of $u$ when $x=b$).
- Substitute into the Integral and Rewrite with New Limits: Rewrite the entire integral in terms of the new variable $u$ and the differential $du$, using the new limits of integration found in Step 2. The integral becomes $\int_{g(a)}^{g(b)} f(u) du$. (Note: The integrand $f(g(x)) g'(x)$ should transform completely into a function of $u$ times $du$).
- Evaluate the Transformed Integral using FTC2: Find an antiderivative $F(u)$ for the integrand $f(u)$ (the function of $u$ in the new integral). Then, evaluate the definite integral using FTC2 with the new limits:
$\int_{g(a)}^{g(b)} f(u) du = [F(u)]_{g(a)}^{g(b)} = F(g(b)) - F(g(a))$.
Crucially, because the limits were changed, you do **not** substitute $u=g(x)$ back into $F(u)$ before applying the limits. You evaluate $F$ at the numerical values $g(a)$ and $g(b)$.
This method is generally faster than substituting back to $x$ before evaluation, provided the transformation of the limits is straightforward.
Example 1. Evaluate $\int_{0}^{1} x (x^2+1)^3 dx$.
Answer:
We need to evaluate the definite integral $\int_{0}^{1} x (x^2+1)^3 dx$. The integrand is a product involving a composite function $(x^2+1)^3$ and a factor $x$ which is related to the derivative of the inner part $x^2+1$. This suggests using substitution.
Step 1: Choose Substitution and find differential.
Let $u$ be the inner function:
"$u = x^2+1$"
Find the differential $du$:
"$du = \frac{d}{dx}(x^2+1) dx = 2x dx$"
From this, we have $x dx = \frac{1}{2} du$.
Step 2: Change the Limits of Integration.
The original limits for $x$ are $a=0$ and $b=1$. We need to find the corresponding values for $u$ using the substitution $u = x^2+1$.
- Lower Limit for $u$: When $x=0$, $u = (0)^2 + 1 = 0 + 1 = 1$. The new lower limit is 1.
- Upper Limit for $u$: When $x=1$, $u = (1)^2 + 1 = 1 + 1 = 2$. The new upper limit is 2.
Step 3: Substitute into the Integral and Rewrite with New Limits.
The original integral is $\int_{0}^{1} (x^2+1)^3 (x dx)$. Substitute $u = x^2+1$ and $x dx = \frac{1}{2} du$. Use the new limits from 1 to 2.
"$\int_{0}^{1} x (x^2+1)^3 dx = \int_{1}^{2} u^3 \left(\frac{1}{2} du\right)$"
[Substitute $u$, $x dx$, and limits]
Move the constant outside:
$= \frac{1}{2} \int_{1}^{2} u^3 du$"
[Constant Multiple Rule]
The integral is now entirely in terms of $u$ with the new limits. We can evaluate this directly using FTC2.
Step 4: Evaluate the Transformed Integral using FTC2.
Find an antiderivative of $f(u) = u^3$. Using the Power Rule for Integration, $\int u^3 du = \frac{u^4}{4} + C$. We choose $F(u) = \frac{u^4}{4}$.
Now evaluate $\frac{1}{2} \int_{1}^{2} u^3 du = \frac{1}{2} [F(u)]_{1}^{2}$ using FTC2 and the new limits [1, 2].
"$= \frac{1}{2} \left[ \frac{u^4}{4} \right]_{1}^{2}$"
[Evaluate using FTC2]
Evaluate at the upper limit ($u=2$) and subtract the evaluation at the lower limit ($u=1$).
"$= \frac{1}{2} \left[ \left( \frac{2^4}{4} \right) - \left( \frac{1^4}{4} \right) \right]$"
[Evaluate at limits]
$= \frac{1}{2} \left[ \frac{16}{4} - \frac{1}{4} \right]$"
[Simplify powers]
$= \frac{1}{2} \left[ 4 - \frac{1}{4} \right]$"
Combine the terms inside the bracket:
$= \frac{1}{2} \left[ \frac{16}{4} - \frac{1}{4} \right] = \frac{1}{2} \left[ \frac{15}{4} \right]$"
Multiply the fractions:
$= \frac{15}{8}$"
The value of the definite integral is $\frac{15}{8}$.
Alternative Method (Method 1: Substitute back to $x$):
First, find the indefinite integral $\int x(x^2+1)^3 dx$. Using the substitution $u=x^2+1, du=2x dx$, this integral is $\int u^3 (\frac{1}{2} du) = \frac{1}{2} \int u^3 du = \frac{1}{2} \frac{u^4}{4} + C = \frac{(x^2+1)^4}{8} + C$. Let $F(x) = \frac{(x^2+1)^4}{8}$.
Now evaluate using the original limits $a=0, b=1$ and FTC2:
"$\int_{0}^{1} x (x^2+1)^3 dx = \left[ \frac{(x^2+1)^4}{8} \right]_{0}^{1}$"
"$= \frac{((1)^2+1)^4}{8} - \frac{((0)^2+1)^4}{8}$"
"$= \frac{(1+1)^4}{8} - \frac{(0+1)^4}{8} = \frac{2^4}{8} - \frac{1^4}{8}$"
"$= \frac{16}{8} - \frac{1}{8} = \frac{15}{8}$"
Both methods give the same result. Changing the limits is generally faster when evaluating definite integrals by substitution.
Properties of Definite Integrals and their Application in Evaluation
Definite integrals have several useful properties that can simplify their evaluation and help in understanding their geometric and analytical characteristics. These properties are derived from the definition of the definite integral as a limit of Riemann sums and the properties of limits and sums, or from the Fundamental Theorems of Calculus.
Let $f$ and $g$ be functions that are integrable on the relevant intervals, and let $k$ be a real number constant.
Basic Properties
These properties often relate to the limits of integration and the combination of integrands.
-
Integral over Zero Width: If the upper and lower limits of integration are the same, the definite integral is zero.
$\int_{a}^{a} f(x) dx = 0$
Reason: Geometrically, this represents the area of a region with zero width. Analytically, using FTC2, $\int_{a}^{a} f(x) dx = F(a) - F(a) = 0$.
-
Reversing Limits of Integration: If the upper and lower limits of integration are swapped, the value of the integral changes sign.
$\int_{b}^{a} f(x) dx = - \int_{a}^{b} f(x) dx$
Reason: Using FTC2, $\int_{b}^{a} f(x) dx = F(a) - F(b) = -(F(b) - F(a)) = - \int_{a}^{b} f(x) dx$.
-
Constant Multiple Property: A constant factor within the integrand can be moved outside the integral sign.
$\int_{a}^{b} k f(x) dx = k \int_{a}^{b} f(x) dx$
Reason: Follows from the constant multiple property of Riemann sums and the linearity of limits, or from the constant multiple rule for differentiation applied to the antiderivative in FTC2: $\int_{a}^{b} k f(x) dx = [k F(x)]_{a}^{b} = k F(b) - k F(a) = k(F(b) - F(a)) = k \int_{a}^{b} f(x) dx$.
-
Sum/Difference Property (Linearity): The integral of a sum or difference of functions is the sum or difference of their individual integrals.
$\int_{a}^{b} [f(x) \pm g(x)] dx = \int_{a}^{b} f(x) dx \pm \int_{a}^{b} g(x) dx$
Reason: Follows from the sum/difference property of Riemann sums and the linearity of limits, or from the sum/difference rule for differentiation applied to the antiderivative in FTC2.
-
Additivity of Intervals: If $a, b,$ and $c$ are three points such that $f$ is integrable on the intervals between them, then the integral from $a$ to $b$ can be split into two integrals with an intermediate point $c$.
$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$
This property holds regardless of the order of $a, b,$ and $c$, as long as $f$ is integrable on the intervals involved. Geometrically, if $a
Reason: Using FTC2, $\int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx = (F(c) - F(a)) + (F(b) - F(c)) = F(c) - F(a) + F(b) - F(c) = F(b) - F(a) = \int_{a}^{b} f(x) dx$.
Properties Related to Symmetry
These properties apply when the interval of integration is symmetric about the y-axis (i.e., of the form $[-a, a]$) and the integrand has even or odd symmetry.
-
Integral of an Even Function over a Symmetric Interval: If $f(x)$ is an even function (i.e., $f(-x) = f(x)$ for all $x$ in the domain) and is integrable on $[-a, a]$, then:
$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$
Reason: Geometrically, for an even function, the area under the curve from $-a$ to 0 is equal to the area from 0 to $a$. Analytically, this can be proven using substitution ($u=-x$) and the additivity property.
-
Integral of an Odd Function over a Symmetric Interval: If $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x$ in the domain) and is integrable on $[-a, a]$, then:
$\int_{-a}^{a} f(x) dx = 0$
Reason: Geometrically, for an odd function, the signed area from $-a$ to 0 is the negative of the signed area from 0 to $a$. The area below the x-axis on one side cancels out the equal area above the x-axis on the other side. Analytically, this is proven using substitution ($u=-x$) and the additivity property.
Other Useful Properties
These properties often involve manipulating the integrand or the limits of integration through substitution or reflection.
-
Variable Substitution Property (Dummy Variable Property): The name of the variable of integration does not affect the value of the definite integral. It is a "dummy" variable.
$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt = \int_{a}^{b} f(u) du$
Reason: FTC2 gives $[F(x)]_{a}^{b} = F(b) - F(a)$ and $[F(t)]_{a}^{b} = F(b) - F(a)$.
-
Reflection Property about the Midpoint of $[0, a]$: For an integral from 0 to $a$, substituting $x$ with $(a-x)$ in the integrand does not change the value of the integral.
$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$
Proof using substitution: Let $u = a-x$. Then $du = -dx$, so $dx = -du$. When $x=0$, $u = a-0=a$. When $x=a$, $u = a-a=0$. The integral becomes $\int_{a}^{0} f(a-u) (-du) = - \int_{a}^{0} f(a-u) du$. Using property 2 ($\int_{b}^{a} = - \int_{a}^{b}$), $- \int_{a}^{0} f(a-u) du = - (-\int_{0}^{a} f(a-u) du) = \int_{0}^{a} f(a-u) du$. Since the variable name does not matter (property 8), $\int_{0}^{a} f(a-u) du = \int_{0}^{a} f(a-x) dx$.
-
Reflection Property about the Midpoint of $[a, b]$: Substituting $x$ with $(a+b-x)$ in the integrand does not change the value of the integral.
$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$
Proof using substitution: Let $u = a+b-x$. Then $du = -dx$, so $dx = -du$. When $x=a$, $u = a+b-a=b$. When $x=b$, $u = a+b-b=a$. The integral becomes $\int_{b}^{a} f(a+b-u) (-du) = - \int_{b}^{a} f(a+b-u) du = \int_{a}^{b} f(a+b-u) du = \int_{a}^{b} f(a+b-x) dx$.
-
Splitting Property for Integral from 0 to $2a$: The integral from 0 to $2a$ can be split into two integrals from 0 to $a$.
$\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{a}^{2a} f(x) dx$
Using substitution $u = 2a-x$ in the second integral $\int_{a}^{2a} f(x) dx$, we get $x=2a-u$, $dx=-du$. When $x=a, u=a$. When $x=2a, u=0$. $\int_{a}^{2a} f(x) dx = \int_{a}^{0} f(2a-u) (-du) = \int_{0}^{a} f(2a-u) du = \int_{0}^{a} f(2a-x) dx$.
So, $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a-x) dx$.
Furthermore, based on the property of $f(2a-x)$:
- If $f(2a-x) = f(x)$ (symmetric about $x=a$), then $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
- If $f(2a-x) = -f(x)$ (anti-symmetric about $x=a$), then $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} (-f(x)) dx = \int_{0}^{a} f(x) dx - \int_{0}^{a} f(x) dx = 0$.
These properties are frequently used to simplify definite integrals before evaluation, especially those involving trigonometric functions or functions with symmetry.
Example 1. Evaluate $\int_{-\pi/2}^{\pi/2} \sin^7 x dx$.
Answer:
We need to evaluate the definite integral $\int_{-\pi/2}^{\pi/2} \sin^7 x dx$. The interval of integration is $[-\pi/2, \pi/2]$, which is symmetric about the origin (of the form $[-a, a]$ where $a = \pi/2$). This suggests checking if the integrand is an even or odd function.
Let $f(x) = \sin^7 x$. To check for even or odd symmetry, we evaluate $f(-x)$.
"$f(-x) = \sin^7 (-x)$"
[Substitute $-x$ for $x$]
Using the property of the sine function $\sin(-x) = -\sin x$:
$= (-\sin x)^7$"
Since the exponent is odd (7), $(-1)^7 = -1$:
$= -(\sin x)^7 = -\sin^7 x$"
So, $f(-x) = -\sin^7 x = -f(x)$.
Since $f(-x) = -f(x)$ for all $x$, the function $f(x) = \sin^7 x$ is an odd function.
Now, we use the property of definite integrals for odd functions over a symmetric interval $[-a, a]$ (Property 7):
"$\int_{-a}^{a} (\text{odd function}) dx = 0$"
Here, $a = \pi/2$. Therefore, the integral is directly 0.
"$\int_{-\pi/2}^{\pi/2} \sin^7 x dx = 0$"
The value of the definite integral is 0.
Example 2. Evaluate $\int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$.
Answer:
We need to evaluate the definite integral $I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$. The limits of integration are 0 and $\pi/2$. The integrand is a rational function of $\sin x$ and $\cos x$. This structure, with limits 0 and $a=\pi/2$, suggests using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ (Property 9).
Let $f(x) = \frac{\sin x}{\sin x + \cos x}$ and $a = \pi/2$. We apply the property:
"$I = \int_{0}^{\pi/2} f(x) dx = \int_{0}^{\pi/2} f(\pi/2 - x) dx$"
Evaluate $f(\pi/2 - x)$ by substituting $(\pi/2 - x)$ for $x$ in the expression for $f(x)$:
"$f(\pi/2 - x) = \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)}$"
Using the complementary angle identities: $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$.
$= \frac{\cos x}{\cos x + \sin x}$"
So, the integral becomes:
"$I = \int_{0}^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx$"
... (i)
Let the original integral be:
"$I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$"
... (ii)
Now, add the two expressions for $I$, equation (i) and equation (ii). We can add the integrands because the limits of integration are the same.
"$I + I = \int_{0}^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx + \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$"
"$2I = \int_{0}^{\pi/2} \left( \frac{\cos x}{\sin x + \cos x} + \frac{\sin x}{\sin x + \cos x} \right) dx$"
[Using Sum Property]
Combine the fractions in the integrand since they have a common denominator:
"$2I = \int_{0}^{\pi/2} \frac{\cos x + \sin x}{\sin x + \cos x} dx$"
The numerator and denominator are the same (assuming $\sin x + \cos x \neq 0$ on the interval, which is true for $x \in [0, \pi/2]$). So the fraction simplifies to 1.
"$2I = \int_{0}^{\pi/2} 1 dx$"
[Simplifying the integrand]
Now, evaluate the simple integral $\int 1 dx = x$. Use FTC2 with the limits 0 and $\pi/2$:
"$2I = [x]_{0}^{\pi/2}$"
"$= (\pi/2) - (0) = \pi/2$"
[Evaluate at limits]
We have $2I = \pi/2$. Solve for $I$ by dividing by 2:
"$I = \frac{\pi/2}{2} = \frac{\pi}{4}$"
The value of the definite integral is $\frac{\pi}{4}$.