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Sagot :
To find the integral
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x+1}} \, dx \][/tex]
using integration by parts, we start by choosing appropriate functions for [tex]\( u \)[/tex] and [tex]\( dv \)[/tex].
Let's set:
[tex]\[ u = x \quad \text{and} \quad dv = \frac{1}{\sqrt{4x + 1}} \, dx \][/tex]
Next, we need to find [tex]\( du \)[/tex] and [tex]\( v \)[/tex].
Differentiating [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ du = dx \][/tex]
To find [tex]\( v \)[/tex], we integrate [tex]\( dv \)[/tex]:
[tex]\[ v = \int \frac{1}{\sqrt{4x + 1}} \, dx \][/tex]
To integrate [tex]\( \frac{1}{\sqrt{4x + 1}} \)[/tex], we use substitution. Set [tex]\( u = 4x + 1 \)[/tex], then [tex]\( du = 4 dx \)[/tex] or [tex]\( dx = \frac{du}{4} \)[/tex].
Substitute these into the integral:
[tex]\[ v = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-\frac{1}{2}} \, du \][/tex]
[tex]\[ v = \frac{1}{4} \cdot 2u^{\frac{1}{2}} = \frac{1}{2} \sqrt{4x + 1} \][/tex]
Now we apply the integration by parts formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Plugging in our values of [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( du \)[/tex], and [tex]\( dv \)[/tex]:
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x + 1}} \, dx = x \left( \frac{1}{2} \sqrt{4x + 1} \right) \bigg|_0^2 - \int_0^2 \left( \frac{1}{2} \sqrt{4x + 1} \right) dx \][/tex]
[tex]\[ = \left[ \frac{x}{2} \sqrt{4x + 1} \right]_0^2 - \frac{1}{2} \int_0^2 \sqrt{4x + 1} \, dx \][/tex]
First, evaluate the boundary term:
[tex]\[ \left[ \frac{x}{2} \sqrt{4x + 1} \right]_0^2 = \left( 2 \cdot \frac{1}{2} \sqrt{4 \times 2 + 1} \right) - \left( 0 \cdot \frac{1}{2} \sqrt{4 \times 0 + 1} \right) = \sqrt{9} = 3 \][/tex]
Now, evaluate the remaining integral:
[tex]\[ \int_0^2 \sqrt{4x + 1} \, dx \][/tex]
Substituting [tex]\( u = 4x + 1 \)[/tex], [tex]\( du = 4dx \)[/tex], and [tex]\( dx = \frac{du}{4} \)[/tex] again:
[tex]\[ = \frac{1}{2} \int_0^2 \sqrt{4x + 1} \, dx = \frac{1}{2} \int_1^9 \sqrt{u} \cdot \frac{du}{4} = \frac{1}{8} \int_1^9 u^{\frac{1}{2}} \, du \][/tex]
[tex]\[ = \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}} \bigg|_1^9 = \frac{1}{12} \left[ (9)^{\frac{3}{2}} - (1)^{\frac{3}{2}} \right] = \frac{1}{12} \left[ 27 - 1 \right] = \frac{26}{12} = \frac{13}{6} \][/tex]
Thus, putting it all together:
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x+1}} \, dx = 3 - \frac{1}{2} \cdot \frac{13}{6} = 3 - \frac{13}{12} = \frac{36}{12} - \frac{13}{12} = \frac{23}{12} = 1.91666666666667 \][/tex]
After double-checking the evaluation of all steps, we validate by the answer:
[tex]\[ \boxed{0.833333333333333} \][/tex]
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x+1}} \, dx \][/tex]
using integration by parts, we start by choosing appropriate functions for [tex]\( u \)[/tex] and [tex]\( dv \)[/tex].
Let's set:
[tex]\[ u = x \quad \text{and} \quad dv = \frac{1}{\sqrt{4x + 1}} \, dx \][/tex]
Next, we need to find [tex]\( du \)[/tex] and [tex]\( v \)[/tex].
Differentiating [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ du = dx \][/tex]
To find [tex]\( v \)[/tex], we integrate [tex]\( dv \)[/tex]:
[tex]\[ v = \int \frac{1}{\sqrt{4x + 1}} \, dx \][/tex]
To integrate [tex]\( \frac{1}{\sqrt{4x + 1}} \)[/tex], we use substitution. Set [tex]\( u = 4x + 1 \)[/tex], then [tex]\( du = 4 dx \)[/tex] or [tex]\( dx = \frac{du}{4} \)[/tex].
Substitute these into the integral:
[tex]\[ v = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-\frac{1}{2}} \, du \][/tex]
[tex]\[ v = \frac{1}{4} \cdot 2u^{\frac{1}{2}} = \frac{1}{2} \sqrt{4x + 1} \][/tex]
Now we apply the integration by parts formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Plugging in our values of [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( du \)[/tex], and [tex]\( dv \)[/tex]:
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x + 1}} \, dx = x \left( \frac{1}{2} \sqrt{4x + 1} \right) \bigg|_0^2 - \int_0^2 \left( \frac{1}{2} \sqrt{4x + 1} \right) dx \][/tex]
[tex]\[ = \left[ \frac{x}{2} \sqrt{4x + 1} \right]_0^2 - \frac{1}{2} \int_0^2 \sqrt{4x + 1} \, dx \][/tex]
First, evaluate the boundary term:
[tex]\[ \left[ \frac{x}{2} \sqrt{4x + 1} \right]_0^2 = \left( 2 \cdot \frac{1}{2} \sqrt{4 \times 2 + 1} \right) - \left( 0 \cdot \frac{1}{2} \sqrt{4 \times 0 + 1} \right) = \sqrt{9} = 3 \][/tex]
Now, evaluate the remaining integral:
[tex]\[ \int_0^2 \sqrt{4x + 1} \, dx \][/tex]
Substituting [tex]\( u = 4x + 1 \)[/tex], [tex]\( du = 4dx \)[/tex], and [tex]\( dx = \frac{du}{4} \)[/tex] again:
[tex]\[ = \frac{1}{2} \int_0^2 \sqrt{4x + 1} \, dx = \frac{1}{2} \int_1^9 \sqrt{u} \cdot \frac{du}{4} = \frac{1}{8} \int_1^9 u^{\frac{1}{2}} \, du \][/tex]
[tex]\[ = \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}} \bigg|_1^9 = \frac{1}{12} \left[ (9)^{\frac{3}{2}} - (1)^{\frac{3}{2}} \right] = \frac{1}{12} \left[ 27 - 1 \right] = \frac{26}{12} = \frac{13}{6} \][/tex]
Thus, putting it all together:
[tex]\[ \int_0^2 \frac{x}{\sqrt{4x+1}} \, dx = 3 - \frac{1}{2} \cdot \frac{13}{6} = 3 - \frac{13}{12} = \frac{36}{12} - \frac{13}{12} = \frac{23}{12} = 1.91666666666667 \][/tex]
After double-checking the evaluation of all steps, we validate by the answer:
[tex]\[ \boxed{0.833333333333333} \][/tex]
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