Math / Calculus

QuestionUsing the substitution: $u=x-3$ . Re-write the indefinite integral then evaluate in terms of $u$ . $\int \frac{x^{2}}{\sqrt{x-3}} d x=\int \square=\square$
Note: answer should be in terms of $u$ only

Studdy Solution

STEP 1

What is this asking? We're taking an integral that's tricky in terms of $x$ and making it easier by rewriting it in terms of $u$ , where $u = x - 3$ , and then solving it. Watch out! Don't forget to substitute back in for $x$ in terms of $u$ everywhere $x$ appears, including $\mathrm{d}x$ !

STEP 2

1. Define the substitution
2. Rewrite the integral
3. Expand and separate the integral
4. Solve each integral part
5. Combine the results and add the constant of integration

STEP 3

We define our substitution as $u = x - 3$ .
This means $x = u + 3$ .

STEP 4

We also need to find $\mathrm{d}x$ in terms of $\mathrm{d}u$ .
Since $x = u + 3$ , we can differentiate both sides with respect to $x$ to get $\frac{\mathrm{d}x}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{\mathrm{d}}{\mathrm{d}x} 3$ , which simplifies to $1 = \frac{\mathrm{d}u}{\mathrm{d}x}$ , so $\mathrm{d}x = \mathrm{d}u$ .

STEP 5

Let's rewrite the integral using our substitution: $\int \frac{x^2}{\sqrt{x-3}} \mathrm{d}x = \int \frac{(u+3)^2}{\sqrt{u}} \mathrm{d}u$

STEP 6

Expand the numerator: $(u+3)^2 = u^2 + 6u + 9$ .

STEP 7

Rewrite the integral as a sum of simpler integrals: $\int \frac{u^2 + 6u + 9}{\sqrt{u}} \mathrm{d}u = \int \frac{u^2}{u^{1/2}} \mathrm{d}u + \int \frac{6u}{u^{1/2}} \mathrm{d}u + \int \frac{9}{u^{1/2}} \mathrm{d}u$

STEP 8

Simplify the exponents: $\int u^{3/2} \mathrm{d}u + \int 6u^{1/2} \mathrm{d}u + \int 9u^{-1/2} \mathrm{d}u$

STEP 9

Apply the power rule of integration: $\int u^n \mathrm{d}u = \frac{u^{n+1}}{n+1} + C$ .

STEP 10

Calculate each integral: $\int u^{3/2} \mathrm{d}u = \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}$ $\int 6u^{1/2} \mathrm{d}u = 6 \cdot \frac{u^{3/2}}{3/2} = 6 \cdot \frac{2}{3}u^{3/2} = 4u^{3/2}$ $\int 9u^{-1/2} \mathrm{d}u = 9 \cdot \frac{u^{1/2}}{1/2} = 9 \cdot 2u^{1/2} = 18u^{1/2}$

STEP 11

Put it all together: $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$

STEP 12

The final answer is $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$ .

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QuestionUsing the substitution: $u=x-3$ . Re-write the indefinite integral then evaluate in terms of $u$ . $\int \frac{x^{2}}{\sqrt{x-3}} d x=\int \square=\square$
Note: answer should be in terms of $u$ only

Studdy Solution

STEP 1

What is this asking? We're taking an integral that's tricky in terms of $x$ and making it easier by rewriting it in terms of $u$ , where $u = x - 3$ , and then solving it. Watch out! Don't forget to substitute back in for $x$ in terms of $u$ everywhere $x$ appears, including $\mathrm{d}x$ !

STEP 2

1. Define the substitution
2. Rewrite the integral
3. Expand and separate the integral
4. Solve each integral part
5. Combine the results and add the constant of integration

STEP 3

We define our substitution as $u = x - 3$ .
This means $x = u + 3$ .

STEP 4

STEP 5

Let's rewrite the integral using our substitution: $\int \frac{x^2}{\sqrt{x-3}} \mathrm{d}x = \int \frac{(u+3)^2}{\sqrt{u}} \mathrm{d}u$

STEP 6

Expand the numerator: $(u+3)^2 = u^2 + 6u + 9$ .

STEP 7

Rewrite the integral as a sum of simpler integrals: $\int \frac{u^2 + 6u + 9}{\sqrt{u}} \mathrm{d}u = \int \frac{u^2}{u^{1/2}} \mathrm{d}u + \int \frac{6u}{u^{1/2}} \mathrm{d}u + \int \frac{9}{u^{1/2}} \mathrm{d}u$

STEP 8

Simplify the exponents: $\int u^{3/2} \mathrm{d}u + \int 6u^{1/2} \mathrm{d}u + \int 9u^{-1/2} \mathrm{d}u$

STEP 9

Apply the power rule of integration: $\int u^n \mathrm{d}u = \frac{u^{n+1}}{n+1} + C$ .

STEP 10

Calculate each integral: $\int u^{3/2} \mathrm{d}u = \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}$ $\int 6u^{1/2} \mathrm{d}u = 6 \cdot \frac{u^{3/2}}{3/2} = 6 \cdot \frac{2}{3}u^{3/2} = 4u^{3/2}$ $\int 9u^{-1/2} \mathrm{d}u = 9 \cdot \frac{u^{1/2}}{1/2} = 9 \cdot 2u^{1/2} = 18u^{1/2}$

STEP 11

Put it all together: $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$

STEP 12

The final answer is $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$ .

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QuestionUsing the substitution: u=x−3u=x-3u=x−3. Re-write the indefinite integral then evaluate in terms of uuu. ∫x2x−3dx=∫□=□\int \frac{x^{2}}{\sqrt{x-3}} d x=\int \square=\square∫x−3​x2​dx=∫□=□Note: answer should be in terms of uuu only

Studdy Solution

STEP 1

STEP 2

1. Define the substitution2. Rewrite the integral3. Expand and separate the integral4. Solve each integral part5. Combine the results and add the constant of integration

STEP 3

We **define** our substitution as u=x−3u = x - 3u=x−3.This means x=u+3x = u + 3x=u+3.

STEP 4

STEP 5

Let's **rewrite** the integral using our substitution: ∫x2x−3dx=∫(u+3)2udu \int \frac{x^2}{\sqrt{x-3}} \mathrm{d}x = \int \frac{(u+3)^2}{\sqrt{u}} \mathrm{d}u ∫x−3​x2​dx=∫u​(u+3)2​du

STEP 6

**Expand** the numerator: (u+3)2=u2+6u+9(u+3)^2 = u^2 + 6u + 9(u+3)2=u2+6u+9.

STEP 7

STEP 8

**Simplify** the exponents: ∫u3/2du+∫6u1/2du+∫9u−1/2du \int u^{3/2} \mathrm{d}u + \int 6u^{1/2} \mathrm{d}u + \int 9u^{-1/2} \mathrm{d}u ∫u3/2du+∫6u1/2du+∫9u−1/2du

STEP 9

**Apply** the power rule of integration: ∫undu=un+1n+1+C\int u^n \mathrm{d}u = \frac{u^{n+1}}{n+1} + C∫undu=n+1un+1​+C.

STEP 10

STEP 11

**Put it all together**: 25u5/2+4u3/2+18u1/2+C \frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C 52​u5/2+4u3/2+18u1/2+C

STEP 12

The **final answer** is 25u5/2+4u3/2+18u1/2+C\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C52​u5/2+4u3/2+18u1/2+C.

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QuestionUsing the substitution: $u=x-3$ . Re-write the indefinite integral then evaluate in terms of $u$ . $\int \frac{x^{2}}{\sqrt{x-3}} d x=\int \square=\square$
Note: answer should be in terms of $u$ only

1. Define the substitution
2. Rewrite the integral
3. Expand and separate the integral
4. Solve each integral part
5. Combine the results and add the constant of integration

We define our substitution as $u = x - 3$ .
This means $x = u + 3$ .

Let's rewrite the integral using our substitution: $\int \frac{x^2}{\sqrt{x-3}} \mathrm{d}x = \int \frac{(u+3)^2}{\sqrt{u}} \mathrm{d}u$

Expand the numerator: $(u+3)^2 = u^2 + 6u + 9$ .

Simplify the exponents: $\int u^{3/2} \mathrm{d}u + \int 6u^{1/2} \mathrm{d}u + \int 9u^{-1/2} \mathrm{d}u$

Apply the power rule of integration: $\int u^n \mathrm{d}u = \frac{u^{n+1}}{n+1} + C$ .

Put it all together: $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$

The final answer is $\frac{2}{5}u^{5/2} + 4u^{3/2} + 18u^{1/2} + C$ .