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Math  /  Calculus

QuestionFind an antiderivative for each function when C=0\mathrm{C}=0. a. 43x3\frac{4}{3} \sqrt[3]{x} b. 14x34\frac{1}{4 \sqrt[4]{x^{3}}} c. x4+1x4\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}}

Studdy Solution

STEP 1

What is this asking? We need to find the antiderivative of each of these functions, assuming the constant of integration is zero. Watch out! Remember the power rule for integration, and don't forget to rewrite radicals as fractional exponents!

STEP 2

1. Antiderivative of 43x3 \frac{4}{3} \sqrt[3]{x}
2. Antiderivative of 14x34 \frac{1}{4 \sqrt[4]{x^{3}}}
3. Antiderivative of x4+1x4 \sqrt[4]{x}+\frac{1}{\sqrt[4]{x}}

STEP 3

Let's **rewrite** the function using fractional exponents: 43x3=43x13 \frac{4}{3} \sqrt[3]{x} = \frac{4}{3} x^{\frac{1}{3}} .
This makes it easier to apply the power rule for integration!

STEP 4

**Apply the power rule**: Remember, the power rule for integration says ∫xn dx=xn+1n+1+C \int x^n \, dx = \frac{x^{n+1}}{n+1} + C , where *C* is the constant of integration.
In our case, *C* = 0.
So, we have ∫43x13 dx=43⋅x13+113+1 \int \frac{4}{3} x^{\frac{1}{3}} \, dx = \frac{4}{3} \cdot \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} .

STEP 5

**Simplify the exponent**: 13+1=13+33=43 \frac{1}{3} + 1 = \frac{1}{3} + \frac{3}{3} = \frac{4}{3} .
So, our antiderivative becomes 43⋅x4343 \frac{4}{3} \cdot \frac{x^{\frac{4}{3}}}{\frac{4}{3}} .

STEP 6

**Divide to one**: Notice we have a 43 \frac{4}{3} in the numerator and denominator!
Dividing them gives us **1**.
Our **final antiderivative** is x43 x^{\frac{4}{3}} .

STEP 7

**Rewrite with fractional exponents**: 14x34=14x−34 \frac{1}{4 \sqrt[4]{x^{3}}} = \frac{1}{4} x^{-\frac{3}{4}} .
Writing it this way makes it much easier to use the power rule!

STEP 8

**Apply the power rule**: ∫14x−34 dx=14⋅x−34+1−34+1 \int \frac{1}{4} x^{-\frac{3}{4}} \, dx = \frac{1}{4} \cdot \frac{x^{-\frac{3}{4} + 1}}{-\frac{3}{4} + 1} .

STEP 9

**Simplify the exponent**: −34+1=−34+44=14 -\frac{3}{4} + 1 = -\frac{3}{4} + \frac{4}{4} = \frac{1}{4} .
This gives us 14⋅x1414 \frac{1}{4} \cdot \frac{x^{\frac{1}{4}}}{\frac{1}{4}} .

STEP 10

**Divide to one**: We have a 14 \frac{1}{4} on the top and bottom, and 14 \frac{1}{4} divided by 14 \frac{1}{4} is **1**!
Our **antiderivative** is x14 x^{\frac{1}{4}} .

STEP 11

**Rewrite with fractional exponents**: x4+1x4=x14+x−14 \sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = x^{\frac{1}{4}} + x^{-\frac{1}{4}} .
This sets us up perfectly to apply the power rule!

STEP 12

**Apply the power rule to each term**: ∫(x14+x−14) dx=x14+114+1+x−14+1−14+1 \int (x^{\frac{1}{4}} + x^{-\frac{1}{4}}) \, dx = \frac{x^{\frac{1}{4} + 1}}{\frac{1}{4} + 1} + \frac{x^{-\frac{1}{4} + 1}}{-\frac{1}{4} + 1} .

STEP 13

**Simplify the exponents**: 14+1=14+44=54 \frac{1}{4} + 1 = \frac{1}{4} + \frac{4}{4} = \frac{5}{4} and −14+1=−14+44=34 -\frac{1}{4} + 1 = -\frac{1}{4} + \frac{4}{4} = \frac{3}{4} .
This gives us x5454+x3434 \frac{x^{\frac{5}{4}}}{\frac{5}{4}} + \frac{x^{\frac{3}{4}}}{\frac{3}{4}} .

STEP 14

**Rewrite the fractions**: Remember dividing by a fraction is the same as multiplying by its reciprocal!
So, we get 45x54+43x34 \frac{4}{5}x^{\frac{5}{4}} + \frac{4}{3}x^{\frac{3}{4}} as our **final antiderivative**.

STEP 15

a. x43 x^{\frac{4}{3}} b. x14 x^{\frac{1}{4}} c. 45x54+43x34 \frac{4}{5}x^{\frac{5}{4}} + \frac{4}{3}x^{\frac{3}{4}}

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