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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 208 to 225.
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Question : 215 of 271
Marks: +1, -0
∫02x2−xdx\int\limits_{0}^{2} x \sqrt{2-x} dx
Solution:  
Let ∫02x2−xdx\int\limits_{0}^{2} x \sqrt{2-x} dx
Put 2−x=t2 - x = t ⇒−−dx=dt\Rightarrow - -dx = dt ; When x=0,  t=2x = 0,\; t = 2 and when x=2,  t=0x = 2,\; t = 0
∴I=−∫20(2−t)tdt\therefore I = -\int\limits_{2}^{0} (2-t) \sqrt{t} dt
=−∫20(2t1/2−t3/2)dt= -\int\limits_{2}^{0} (2t^{1/2} - t^{3/2}) dt
=[2t3/23/2−t5/25/2]02= \left[ \frac{2t^{3/2}}{3/2} - \frac{t^{5/2}}{5/2} \right]^2_0 [∵−∫a0f(x)dx=∫0af(x)dx]\left[ \because -\int\limits_{a}^{0} f(x) dx = \int\limits_{0}^{a} f(x) dx \right]
=[43t3/2−25t5/2]02= \left[ \frac{4}{3} t^{3/2} - \frac{2}{5} t^{5/2} \right]^2_0
=43(2)3/2−25(2)5/2= \frac{4}{3}(2)^{3/2} - \frac{2}{5}(2)^{5/2}
=16215= \frac{16\sqrt{2}}{15}
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