JEE Main 2025 3 April Morning Shift
Exam JEE Main
Year 2025
Date 3 April
Shift Morning Shift
Subject Mathematics
Chapter Indefinite Integration JEE Main 2025 · 3 April · Morning Shift · Mathematics · Indefinite Integration
Let f(x)=∫ x^3 3-x^2 d x . If 5 f( 2 )=-4 , then f(1) is equal to
Let
f ( x ) = ∫ x 3 3 − x 2 d x f(x)=\int x^3 \sqrt{3-x^2} d x f ( x ) = ∫ x 3 3 − x 2 d x . If
5 f ( 2 ) = − 4 5 f(\sqrt{2})=-4 5 f ( 2 ) = − 4 , then
f ( 1 ) f(1) f ( 1 ) is equal to
Options A − 2 2 5 -\frac{2 \sqrt{2}}{5} − 5 2 2 B − 8 2 5 -\frac{8 \sqrt{2}}{5} − 5 8 2 C − 4 2 5 -\frac{4 \sqrt{2}}{5} − 5 4 2 D − 6 2 5 -\frac{6 \sqrt{2}}{5} − 5 6 2 Correct answer D. − 6 2 5 -\frac{6 \sqrt{2}}{5} − 5 6 2
Step-by-step solution
Let 3 − x 2 = t 2 + x d x = − t d t f ( x ) = ∫ ( 3 − t 2 ) ⋅ t ( − t d t ) + c = ∫ ( t 4 − 3 t 2 ) d t + c = t 5 5 − t 3 + c f ( x ) = ( 3 − x 2 ) 5 / 2 5 − ( 3 − x 2 ) 3 / 2 + c f ( 2 ) = 1 5 − 1 + c = − 4 5 \begin{aligned} & \text{ Let } 3-x^2=t^2 \\ & +x d x=-t d t \\ & f(x)=\int\left(3-t^2\right) \cdot t(-t d t)+c \\ & =\int\left(\mathrm{t}^4-3 \mathrm{t}^2\right) \mathrm{dt}+\mathrm{c} \\ & =\frac{\mathrm{t}^5}{5}-\mathrm{t}^3+\mathrm{c} \\ & \mathrm{f}(\mathrm{x})=\frac{\left(3-\mathrm{x}^2\right)^{5 / 2}}{5}-\left(3-\mathrm{x}^2\right)^{3 / 2}+\mathrm{c} \\ & \mathrm{f}(\sqrt{2})=\frac{1}{5}-1+\mathrm{c}=-\frac{4}{5}\end{aligned} Let 3 − x 2 = t 2 + x d x = − t d t f ( x ) = ∫ ( 3 − t 2 ) ⋅ t ( − t d t ) + c = ∫ ( t 4 − 3 t 2 ) dt + c = 5 t 5 − t 3 + c f ( x ) = 5 ( 3 − x 2 ) 5/2 − ( 3 − x 2 ) 3/2 + c f ( 2 ) = 5 1 − 1 + c = − 5 4 c = 0 f ( 1 ) = 2 5 / 2 5 − 2 3 / 2 = 2 1 / 2 ( 4 5 − 2 ) f ( 1 ) − − 6 2 5 \begin{aligned} & \mathrm{c}=0 \\ & \mathrm{f}(1)=\frac{2^{5 / 2}}{5}-2^{3 / 2} \\ & =2^{1 / 2}\left(\frac{4}{5}-2\right) \\ & \mathrm{f}(1)-\frac{-6 \sqrt{2}}{5}\end{aligned} c = 0 f ( 1 ) = 5 2 5/2 − 2 3/2 = 2 1/2 ( 5 4 − 2 ) f ( 1 ) − 5 − 6 2 $
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← Previous question If ∫ ( 1+x^2 +x)^ 10 ( 1+x^2 -x)^9 d x= (1)/(m)(( 1+x^2 +x)^n(n 1+x^2 -x))+C where C is…Next question → If f(x)=∫ 1 x^ 1 / 4 (1+x^ 1 / 4 ) d x, f(0)=-6 , then f(1) is equal to :
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