The value of -13x2dx2x2-16-8x is

Let nbsp; I=∫_1^3x^2dx2x^2+16 8x ⇒ I=∫_1^3 nbsp;x^2dx(4 x)^2+x^2 ⋯(i) Using property ∫_a^b f(x) d x=∫_a^b f(a+b x) d x We get, I=∫_1^3 nbsp;(4 x)^2dx(4 ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of $3 \int 1 \frac{x^{2} d x}{2 x^{2} + 16 - 8 x}$ is

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Solution

Let
$I = 3 \int 1 \frac{x^{2} d x}{2 x^{2} + 16 - 8 x} \Rightarrow I = 3 \int 1 \frac{x^{2} d x}{(4 - x)^{2} + x^{2}} \dots (i)$
Using property
$b \int a f (x) d x = b \int a f (a + b - x) d x$
We get, $I = 3 \int 1 \frac{(4 - x)^{2} d x}{(4 - x)^{2} + x^{2}} \dots (i i)$
On adding $(i)$ and $(i i),$ we have
$2 I = 3 \int 1 1 d x = 2 ∴ I = 1$

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First Fundamental Theorem of Calculus

Standard XII Mathematics

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The value of 3∫1x2dx2x2+16−8x is

Let I=3∫1x2dx2x2+16−8x⇒I=3∫1x2dx(4−x)2+x2 ⋯(i) Using property b∫af(x)dx=b∫af(a+b−x)dx We get, I=3∫1(4−x)2dx(4−x)2+x2 ⋯(ii) On adding (i) and (ii), we have 2I=3∫11dx=2∴I=1

The value of $3 \int 1 \frac{x^{2} d x}{2 x^{2} + 16 - 8 x}$ is