If `int_(0)^(1)x^(11)e^(-x^(24))dx=A`, and `int_(0)^(1)x^(3)e^(-x^(8))dx=B`, then the relation between A and B is

Options

A. `A = 3B`
B. `B = 3A`
C. `A+3B=0`
D. `B+3A=0`

Answer

To solve the problem, we need to find the relationship between the two integrals given:

1. \( A = \int_{0}^{1} x^{11} e^{-x^{24}} \, dx \)
2. \( B = \int_{0}^{1} x^{3} e^{-x^{8}} \, dx \)

### Step-by-step Solution:

**Step 1: Rewrite the integral for A.**

We start with the integral for \( A \):
\[ A = \int_{0}^{1} x^{11} e^{-x^{24}} \, dx \]

**Step 2: Change of variables.**

Let's make a substitution. Set \( t = x^3 \). Then, we differentiate:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Since \( x = t^{1/3} \), we have:
\[ x^2 = (t^{1/3})^2 = t^{2/3} \]
Thus:
\[ dx = \frac{dt}{3t^{2/3}} \]

**Step 3: Change the limits of integration.**

When \( x = 0 \), \( t = 0^3 = 0 \). When \( x = 1 \), \( t = 1^3 = 1 \). Therefore, the limits remain from 0 to 1.

**Step 4: Substitute into the integral.**

Now substitute \( x \) and \( dx \) into the integral for \( A \):
\[ A = \int_{0}^{1} (t^{1/3})^{11} e^{-(t^{1/3})^{24}} \cdot \frac{dt}{3t^{2/3}} \]
This simplifies to:
\[ A = \int_{0}^{1} \frac{t^{11/3} e^{-t^8}}{3t^{2/3}} \, dt = \frac{1}{3} \int_{0}^{1} t^{11/3 - 2/3} e^{-t^8} \, dt = \frac{1}{3} \int_{0}^{1} t^{9/3} e^{-t^8} \, dt \]
Thus:
\[ A = \frac{1}{3} \int_{0}^{1} t^{3} e^{-t^8} \, dt \]

**Step 5: Recognize the integral.**

Notice that:
\[ \int_{0}^{1} t^{3} e^{-t^8} \, dt = B \]
Therefore:
\[ A = \frac{1}{3} B \]

**Step 6: Express the relationship between A and B.**

From the above, we can express \( B \) in terms of \( A \):
\[ B = 3A \]

### Conclusion:

Thus, the relationship between \( A \) and \( B \) is:
\[ B = 3A \]