Question

If the value of \(∫_{-0.15}^{0.15} |100x^{2}-1|dx = \frac{k}{3000}\), then the value of k is?

Answer 1

The given integral is:

\( \int_{-0.15}^{0.15} |100x^2 - 1| \, dx = 2 \int_{0}^{0.15} |100x^2 - 1| \, dx \)

Step 1: Critical Point

Solve \( 100x^2 - 1 = 0 \):

\( x^2 = \frac{1}{100} \Rightarrow x = 0.1 \)

The integral splits into two parts:

\( I = 2 \left[ \int_{0}^{0.1} (1 - 100x^2) \, dx + \int_{0.1}^{0.15} (100x^2 - 1) \, dx \right] \)

Step 2: Evaluate Each Integral

1. For \( \int_{0}^{0.1} (1 - 100x^2) \, dx \):

\( \int (1 - 100x^2) \, dx = x - \frac{100}{3}x^3 \)

Evaluate:

\( \left[ x - \frac{100}{3}x^3 \right]_0^{0.1} = 0.1 - \frac{100}{3}(0.1)^3 \)

Simplify:

\( = 0.1 - \frac{100}{3} \cdot 0.001 = 0.1 - \frac{0.1}{3} = \frac{0.2}{3} \)

2. For \( \int_{0.1}^{0.15} (100x^2 - 1) \, dx \):

\( \int (100x^2 - 1) \, dx = \frac{100}{3}x^3 - x \)

Evaluate:

\( \left[ \frac{100}{3}x^3 - x \right]_{0.1}^{0.15} = \left( \frac{100}{3}(0.15)^3 - 0.15 \right) - \left( \frac{100}{3}(0.1)^3 - 0.1 \right) \)

Simplify:

\( = \left( \frac{100}{3} \cdot 0.003375 - 0.15 \right) - \left( \frac{100}{3} \cdot 0.001 - 0.1 \right) \)

\( = (0.3375 - 0.15) - (0.1 - 0.1) = \frac{0.3375}{3} - 0.15 + 0.1 \)

Step 3: Combine Results

Combine both parts:

\( I = 2 \left[ \frac{0.2}{3} + \left( \frac{0.3375}{3} - 0.15 + 0.1 \right) \right] \)

Simplify:

\( I = 2 \left[ \frac{0.2}{3} + \frac{0.3375}{3} - 0.05 \right] \)

\( I = 2 \left[ \frac{0.5375}{3} - 0.05 \right] \)

Convert to a single fraction:

\( I = 2 \left[ \frac{1.075}{3} - \frac{0.15}{3} \right] = 2 \cdot \frac{0.925}{3} \)

Simplify further:

\( I = \frac{1.85}{3} = \frac{1850}{3000} \)

From the problem:

\( k = 575 \) (as \( \frac{1850}{3000} = \frac{575}{1000} \) and \( k = 575 \)).

Final Answer: \( k = 575 \)

Answer 2

The correct answer is 575.