The value of \(\displaystyle \int_0^1 \frac{a-bx^2}{(a+bx^2)^2} dx\) is :
1/(a + b)
Evaluating a Definite Integral Problem
We are asked to find the value of the definite integral:
\(\displaystyle \int_0^1 \frac{a-bx^2}{(a+bx^2)^2} dx\)
This is a calculus problem involving integration of a rational function within specific limits, from 0 to 1.
Strategy: Recognizing a Derivative Pattern
Sometimes, complex-looking integrands are actually the result of differentiating a simpler function. Let's consider functions involving \(\frac{x}{a+bx^2}\). The derivative of such a function might resemble our integrand.
Calculating a Related Derivative
Let's calculate the derivative of \(\frac{x}{a+bx^2}\) with respect to \(x\). We can use the quotient rule, which states that if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\).
Here, let \(u(x) = x\) and \(v(x) = a+bx^2\). Then \(u'(x) = 1\) and \(v'(x) = 2bx\).
Applying the quotient rule:
\(\displaystyle \frac{d}{dx} \left( \frac{x}{a+bx^2} \right) = \frac{(1)(a+bx^2) - (x)(2bx)}{(a+bx^2)^2}\)
\(\displaystyle = \frac{a+bx^2 - 2bx^2}{(a+bx^2)^2}\)
\(\displaystyle = \frac{a-bx^2}{(a+bx^2)^2}\)
This is exactly the integrand we have in the definite integral problem!
Applying the Fundamental Theorem of Calculus
Since we found that the integrand \(\frac{a-bx^2}{(a+bx^2)^2}\) is the derivative of \(\frac{x}{a+bx^2}\), we can use the Fundamental Theorem of Calculus to evaluate the definite integral.
The theorem states that if \(F'(x) = f(x)\), then \(\int_c^d f(x) dx = [F(x)]_c^d = F(d) - F(c)\).
In our case, \(f(x) = \frac{a-bx^2}{(a+bx^2)^2}\), and we found that \(F(x) = \frac{x}{a+bx^2}\) is its antiderivative. The limits of integration are from \(c=0\) to \(d=1\).
So, the value of the integral is:
\(\displaystyle \int_0^1 \frac{a-bx^2}{(a+bx^2)^2} dx = \left[ \frac{x}{a+bx^2} \right]_0^1\)
Evaluating at the Limits
Now we substitute the upper limit (\(x=1\)) and the lower limit (\(x=0\)) into the antiderivative and subtract:
Value at upper limit (\(x=1\)):
\(\displaystyle \frac{1}{a+b(1)^2} = \frac{1}{a+b}\)
Value at lower limit (\(x=0\)):
\(\displaystyle \frac{0}{a+b(0)^2} = \frac{0}{a+0} = \frac{0}{a}\)
Note: We assume \(a \neq 0\) here for the lower limit evaluation to be meaningful, although the fraction is 0 regardless. The expression \(a+bx^2\) must also be non-zero over the interval [0, 1], particularly \(a+b \neq 0\).
Subtracting the lower limit value from the upper limit value:
\(\displaystyle \left[ \frac{x}{a+bx^2} \right]_0^1 = \frac{1}{a+b} - \frac{0}{a}\)
\(\displaystyle = \frac{1}{a+b} - 0\)
\(\displaystyle = \frac{1}{a+b}\)
Final Answer
The value of the definite integral \(\displaystyle \int_0^1 \frac{a-bx^2}{(a+bx^2)^2} dx\) is \(\frac{1}{a+b}\).
Revision Table: Key Concepts in Definite Integrals
| Concept | Description | Application in this problem |
|---|---|---|
| Definite Integral | Represents the net area under a curve between two limits. | We evaluated the integral over the interval [0, 1]. |
| Antiderivative (Indefinite Integral) | A function \(F(x)\) whose derivative is \(f(x)\), i.e., \(F'(x) = f(x)\). | We found that \(\frac{x}{a+bx^2}\) is the antiderivative of the integrand. |
| Fundamental Theorem of Calculus Part 2 | If \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_a^b f(x) dx = F(b) - F(a)\). | Used to calculate the definite integral by evaluating the antiderivative at the limits. |
| Quotient Rule for Derivatives | Used to find the derivative of a function that is a ratio of two functions: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). | Used to verify that \(\frac{x}{a+bx^2}\) is indeed the antiderivative. |
Additional Information: Recognizing Integrand Structures
This problem highlights a common technique in integration: recognizing if the integrand is the derivative of a standard or easily guessable function. This is often true for integrands that are rational functions where the denominator is raised to a power, especially if the numerator involves terms related to the derivative of the denominator or parts of it.
For example, integrands of the form \(\frac{f'(x)}{f(x)}\) integrate to \(\ln|f(x)| + C\). Integrands of the form \(\frac{f'(x)}{[f(x)]^n}\) can often be solved using a simple substitution \(u = f(x)\).
In this problem, the structure \(\frac{a-bx^2}{(a+bx^2)^2}\) suggested looking at functions whose derivative involves a squared denominator, such as functions of the form \(\frac{g(x)}{a+bx^2}\). Calculating the derivative of \(\frac{x}{a+bx^2}\) proved to be the correct approach.
Developing this skill requires practice with both differentiation and integration, as recognizing patterns in the integrand is key to choosing the right integration technique.
1/(a + b)
Let \( I = \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} \, dx \)
Note that: \( \frac{d}{dx} \left( \frac{x}{a + bx^2} \right) = \frac{a - bx^2}{(a + bx^2)^2} \)
So, \[ I = \left[ \frac{x}{a + bx^2} \right]_0^1 = \frac{1}{a + b} - 0 = \frac{1}{a + b} \]
Answer:
\( \frac{1}{a + b} \)
1/(a + b)
Step 1: Recognize the derivative pattern
We observe that: \[ \frac{d}{dx}\left(\frac{x}{a + b x^2}\right) = \frac{(1)(a+bx^2) - x(2bx)}{(a+bx^2)^2} = \frac{a - b x^2}{(a + b x^2)^2} \] which exactly matches our integrand.
Step 2: Apply the Fundamental Theorem of Calculus
\[ \int \frac{a - b x^2}{(a + b x^2)^2} dx = \frac{x}{a + b x^2} + C \]
Step 3: Evaluate the definite integral
\[ \int_0^1 \frac{a - b x^2}{(a + b x^2)^2} dx = \left.\frac{x}{a + b x^2}\right|_0^1 = \frac{1}{a + b} - 0 = \frac{1}{a + b} \]
Final Answer:
The value of the integral is \[ \boxed{\frac{1}{a + b}} \].
Key Points:
- The solution uses reverse differentiation (anti-derivative)
- Verification of the derivative pattern was crucial
- All terms cancel perfectly leaving a simple final result
Important Questions from Integrals
$$ \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx = $$
\( \int_{0}^{\frac{\pi}{2}} \frac{1 - \cot x}{\cosec x + \cos x} dx = \)
The value of the integral \( \int_{\log_e 2}^{\log_e 3} \frac{e^{2x}- 1}{e^{2x} + 1} dx \) is :
\(\displaystyle \int \frac{\pi}{x^{n+1} - x} dx\)
\(\int_{2}^{3} |2x - 1| \,dx =\)
