If `I_(m)=int_(0)^(oo) e^(-x)x^(n-1)dx, "then" int_(0)^(oo) e^(-lambdax) x^(n-1)dx`

Options

A. `lambdaI_(n)`
B. `(1)/(lambda)I_(n)`
C. `(I_(n))/(lambda^(n))`
D. `lambda^(n)I_(n)`

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{0}^{\infty} e^{-\lambda x} x^{n-1} \, dx \]

given that

\[ I_m = \int_{0}^{\infty} e^{-x} x^{n-1} \, dx. \]

### Step 1: Change of Variables
We will perform a change of variables by letting \( t = \lambda x \). Then, we can express \( x \) in terms of \( t \):

\[ x = \frac{t}{\lambda}. \]

### Step 2: Compute the Differential
Next, we need to compute the differential \( dx \):

\[ dx = \frac{1}{\lambda} dt. \]

### Step 3: Change the Limits of Integration
Now we will change the limits of integration. When \( x = 0 \), \( t = 0 \), and when \( x = \infty \), \( t = \infty \). Thus, the limits remain the same:

\[ \int_{0}^{\infty} \rightarrow \int_{0}^{\infty}. \]

### Step 4: Substitute into the Integral
Substituting \( x \) and \( dx \) into the integral, we have:

\[ I = \int_{0}^{\infty} e^{-\lambda x} x^{n-1} \, dx = \int_{0}^{\infty} e^{-t} \left(\frac{t}{\lambda}\right)^{n-1} \frac{1}{\lambda} dt. \]

### Step 5: Simplify the Integral
Now we can simplify the integral:

\[ I = \int_{0}^{\infty} e^{-t} \frac{t^{n-1}}{\lambda^{n-1}} \cdot \frac{1}{\lambda} dt = \frac{1}{\lambda^n} \int_{0}^{\infty} e^{-t} t^{n-1} dt. \]

### Step 6: Recognize the Integral
The integral

\[ \int_{0}^{\infty} e^{-t} t^{n-1} dt \]

is equal to \( I_m \):

\[ I_m = \int_{0}^{\infty} e^{-t} t^{n-1} dt. \]

### Step 7: Final Expression
Thus, we can write:

\[ I = \frac{I_m}{\lambda^n}. \]

### Conclusion
The value of the integral

\[ \int_{0}^{\infty} e^{-\lambda x} x^{n-1} \, dx = \frac{I_m}{\lambda^n}. \]