Q.42 The solution to the integral1
2
0
2 1y y dy , rounded off to TWO decimal places,
is ________.

The value of the integral \(\int_{0}^{1} 2y\sqrt{1+y^{2}}\,dy\) rounded to two decimal places is 1.22.

Introduction

The integral \(\int_{0}^{1} 2y\sqrt{1+y^{2}}\,dy\) appears frequently in calculus exams and is a classic example of using substitution to simplify a square‑root expression.  Understanding this type of problem strengthens core integration techniques and improves confidence with definite integrals involving algebraic roots.

Step‑by‑step solution

1. Set up the substitution

The integrand contains \(\sqrt{1+y^{2}}\), so a substitution that simplifies this expression is helpful.  Let \(u = 1 + y^{2}\), then \(\frac{du}{dy} = 2y\) so \(du = 2y\,dy\), which matches the factor in the integrand.

With this change of variable, the integral becomes \(\int 2y\sqrt{1+y^{2}}\,dy = \int \sqrt{u}\,du\) before changing the limits.

2. Change the limits

For \(y = 0\), \(u = 1 + 0^{2} = 1\). For \(y = 1\), \(u = 1 + 1^{2} = 2\).

Therefore the definite integral becomes \(\int_{0}^{1} 2y\sqrt{1+y^{2}}\,dy = \int_{1}^{2} \sqrt{u}\,du\).

3. Integrate in the new variable

Write \(\sqrt{u}\) as \(u^{1/2}\) and integrate to obtain \(\int_{1}^{2} u^{1/2}\,du = \left[\frac{2}{3}u^{3/2}\right]_{1}^{2}\).

Evaluating at the bounds gives \(\left[\frac{2}{3}u^{3/2}\right]_{1}^{2} = \frac{2}{3}\left(2^{3/2} – 1^{3/2}\right) = \frac{2}{3}(2\sqrt{2} – 1)\), which is the exact value of the integral.

4. Numerical value and rounding

Using \(\sqrt{2} \approx 1.41421\), we get \(2\sqrt{2} \approx 2.82842\) and hence \(2\sqrt{2} – 1 \approx 1.82842\).

Multiplying by \(\frac{2}{3}\) gives \(\frac{2}{3} \times 1.82842 \approx 1.21895\), which rounds to 1.22 to two decimal places.