14. \[ \int_0^3 \frac{dx}{\sqrt{1 + x}} =\ ? \]






Answer is: option1

2

Solution:

Let \( u = 1 + x \Rightarrow du = dx \)

When \( x = 0 \), \( u = 1 \)
When \( x = 3 \), \( u = 4 \)

Now the integral becomes:

\[ \int_1^4 \frac{1}{\sqrt{u}} \, du = \int_1^4 u^{-\frac{1}{2}} \, du \]

\[ \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} + C \]

Evaluate the definite integral:

\[ 2u^{\frac{1}{2}} \Big|_1^4 = 2(\sqrt{4}) - 2(\sqrt{1}) = 2(2) - 2(1) = 4 - 2 = 2 \]

Correct Answer is (A) ie., 2

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