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IntegrationQuestion and Answers: Page 16

Question Number 203884 Answers: 1 Comments: 0

find ∫_0 ^∞ (x^3 /((1+x)^4 (x+2)^5 ))dx

$f i n d \int_{0}^{\infty} \frac{x^{3}}{{(1 + x)}^{4} {(x + 2)}^{5}} d x$

Question Number 203867 Answers: 1 Comments: 0

find ∫(√((1−x^3 )/(1+x^3 )))dx

$f i n d \int \sqrt{\frac{1 - x^{3}}{1 + x^{3}}} d x$

Question Number 203836 Answers: 2 Comments: 0

Question Number 203772 Answers: 2 Comments: 0

Question Number 203747 Answers: 2 Comments: 0

Question Number 203679 Answers: 0 Comments: 0

Question Number 203564 Answers: 1 Comments: 0

∫_0 ^( ∞) ((sin^( 3) (x))/x^( 2) ) dx= ?

$\int_{0}^{\infty} \frac{{s i n}^{3} (x)}{x^{2}} d x = ?$

Question Number 203385 Answers: 2 Comments: 0

Question Number 203349 Answers: 1 Comments: 0

calculate ∫∫_([0,a]^2 ) e^(−x^2 −y^2 ) dxdy can you find ∫_0 ^a e^(−x^2 ) dx ? a>0

$c a l c u l a t e \int \int_{{[0, a]}^{2}} e^{- x^{2} - y^{2}} d x d y$ $c a n y o u f i n d \int_{0}^{a} e^{- x^{2}} d x ?$ $a > 0$

Question Number 203186 Answers: 1 Comments: 0

f(x)={_(2 x=1) ^(7 x≠1 ) ⇒ ∫_0 ^( 4) f(x)dx=?

$f (x) = {_{2 x = 1}^{7 x \neq 1} \Rightarrow \int_{0}^{4} f (x) d x = ?$

Question Number 203047 Answers: 0 Comments: 0

Question Number 202930 Answers: 1 Comments: 0

Question Number 202882 Answers: 3 Comments: 0

⇃

$⇃ \underset{―}{}$

Question Number 202636 Answers: 0 Comments: 0

((^3 (√4^(5−x) ))/(∫_4 ^6 (x−1)dx)) = (1/2^(2x−1) ) , find the value of x. Solution (4^((5−x)/3) /(∫_4 ^6 ((x^2 /2)−x+k))) = (1/2^(2x−1) ) (2^(2•((5−x)/3)) /(((6^2 /2)−6+k)−((4^2 /2)−4+k))) = (1/2^(2x−1) ) (2^((10−2x)/3) /(((36)/2)−6+k−((16)/2)+4−k)) = (1/2^(2x−1) ) (2^((10−2x)/3) /(18−6−8+4)) = (1/2^(2x−1) ) (2^((10−2x)/3) /8) = (1/2^(2x−1) ) (Cross Multiply) 2^(2x−1) ×2^((10−2x)/3) = 8×1 2^(2x−1) ×2^((10−2x)/3) = 2^3 2^(2x−1+((10−2x)/3)) = 2^3 (Since, the bases are equal. Then, we can equate the exponents) 2x−1+((10−2x)/3) = 3 (Multiply each term by 3) 3(2x)−3(1)+3(((10−2x)/3)) = 3(3) 6x−3+10−2x = 9 (Collect Like Terms) 4x+7 = 9 4x = 9−7 4x = 2 (Divide Both Sides by 4) ((4x)/4) = (2/4) ∴ x = (1/2)

$\frac{^{3} \sqrt{4^{5 - x}}}{\int_{4}^{6} (x - 1) d x} = \frac{1}{2^{2 x - 1}}, find the value of x .$ $\underset{―}{Solution}$ $\frac{4^{\frac{5 - x}{3}}}{\int_{4}^{6} (\frac{x^{2}}{2} - x + k)} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{2 ∙ \frac{5 - x}{3}}}{(\frac{6^{2}}{2} - 6 + k) - (\frac{4^{2}}{2} - 4 + k)} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{\frac{36}{2} - 6 + k - \frac{16}{2} + 4 - k} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{18 - 6 - 8 + 4} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{8} = \frac{1}{2^{2 x - 1}} (Cross Multiply)$ $2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 8 \times 1$ $2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 2^{3}$ $2^{2 x - 1 + \frac{10 - 2 x}{3}} = 2^{3} (Since, the bases are equal . Then, we can equate the exponents)$ $2 x - 1 + \frac{10 - 2 x}{3} = 3 (Multiply each term by 3)$ $3 (2 x) - 3 (1) + 3 (\frac{10 - 2 x}{3}) = 3 (3)$ $6 x - 3 + 10 - 2 x = 9 (Collect Like Terms)$ $4 x + 7 = 9$ $4 x = 9 - 7$ $4 x = 2 (Divide Both Sides by 4)$ $\frac{4 x}{4} = \frac{2}{4}$ $∴ x = \frac{1}{2}$