Question

The solution to the integral \( \int_0^1 2y\sqrt{1 + y^2} \, dy \), rounded off to TWO decimal places, is ............ 
 

Hint:

When solving integrals involving square roots, look for standard formulas or use numerical methods if necessary.

Answer 1

Correct Answer: 1.2 - 1.3

1. Set up the Integral

The integral to be solved is:

$$I = \int_{0}^{1} 2y\sqrt{1+y^2} \, dy$$

2. Use u-Substitution

This integral can be solved using substitution, which is appropriate for expressions where the integrand contains a function and its derivative (or a multiple of its derivative).

Let the expression under the square root be $u$:

$$u = 1 + y^2$$

Find the differential $du$:

$$\frac{du}{dy} = 2y \implies du = 2y \, dy$$

Change the limits of integration from $y$ to $u$:

Lower limit ($y=0$): $u = 1 + (0)^2 = 1$

Upper limit ($y=1$): $u = 1 + (1)^2 = 2$

3. Evaluate the Integral

Substitute $u$ and $du$ into the integral:

$$I = \int_{u_1}^{u_2} \sqrt{u} \, du = \int_{1}^{2} u^{1/2} \, du$$

Now, apply the power rule for integration, $\int u^n \, du = \frac{u^{n+1}}{n+1}$:

$$I = \left[ \frac{u^{1/2 + 1}}{1/2 + 1} \right]_{1}^{2} = \left[ \frac{u^{3/2}}{3/2} \right]_{1}^{2} = \frac{2}{3} \left[ u^{3/2} \right]_{1}^{2}$$

Apply the limits of integration:

$$I = \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right)$$

$$I = \frac{2}{3} \left( 2\sqrt{2} - 1 \right)$$

4. Numerical Calculation and Rounding

Calculate the numerical value:

$$I \approx \frac{2}{3} (2 \times 1.41421356 - 1)$$

$$I \approx \frac{2}{3} (2.82842712 - 1)$$

$$I \approx \frac{2}{3} (1.82842712)$$

$$I \approx 1.2189514$$

The question asks for the solution rounded off to TWO decimal places:

$$I \approx \mathbf{1.22}$$