₀² x{x+2}dx - NCERT Class XII Mathematics Chapter -

NCERT Class XII Mathematics Chapter - - Solutions

Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.

Question : 201 of 271

Marks: +1, -0

\int\limits_{0}^{2} x\sqrt{x+2}dx

Solution:

Let

I=\int\limits_{0}^{2} x\sqrt{x+2}dx

Put

x + 2 = t \Rightarrow dx = dt

When

x = 0, t = 2

and when

x = 2, t = 4

\therefore I=\int\limits_{2}^{4}(t-2)\sqrt{t}dt

=\int\limits_{2}^{4}(t^{3/2}-2t^{1/2})dt

=\left[\frac{2}{5}t^{5/2}-2\times\frac{2}{3}t^{3/2}\right]_{2}^{4}

=\left[\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}\right]_{2}^{4}

=\left[\frac{2}{5}(4)^{5/2}-\frac{4}{3}(4)^{3/2}\right]-\left[\frac{2}{5}(2)^{5/2}-\frac{4}{3}(2)^{3/2}\right]

=\frac{2}{5}\times32-\frac{4}{3}\times8-\frac{8}{5}\sqrt{2}+\frac{8}{3}\sqrt{2}

=\frac{32}{15}+\frac{16\sqrt{2}}{15}

=\frac{16\sqrt{2}}{15}(\sqrt{2}-1)

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