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Q7:
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 7. $\int_{0}^{1} x(1-x)^n dx$

Solution :

Step 1: Application of Definite Integral Property

To evaluate the integral $I = \int_{0}^{1} x(1-x)^n \, dx$, we utilize the property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$. [By the Property of Definite Integrals: King's Property].

Here, $a = 0$ and $b = 1$. Substituting $x$ with $(1 - x)$:

$I = \int_{0}^{1} (1 - x)(1 - (1 - x))^n \, dx$

Step 2: Algebraic Simplification

Simplify the integrand by distributing the terms:

$I = \int_{0}^{1} (1 - x)(x)^n \, dx$ $I = \int_{0}^{1} (x^n - x^{n+1}) \, dx$

Step 3: Integration

Applying the Fundamental Theorem of Calculus, we compute the anti-derivative of the polynomial expression [Per the Power Rule of Integration: $\int x^n dx = \frac{x^{n+1}}{n+1}$]:

$I = \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1}$

Step 4: Evaluating Limits

Substitute the upper and lower limits of integration:

$I = \left( \frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2} \right) - (0 - 0)$ $I = \frac{1}{n+1} - \frac{1}{n+2}$

Step 5: Algebraic Consolidation

Find the common denominator to simplify the result:

$I = \frac{(n+2) - (n+1)}{(n+1)(n+2)}$ $I = \frac{n + 2 - n - 1}{(n+1)(n+2)}$ $I = \frac{1}{(n+1)(n+2)}$ x y 0 1 f(x) = x(1-x)^n

Final Solution: \frac{1}{(n+1)(n+2)}


More Questions from Class 12 Mathematics Integrals EXERCISE 7.10


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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