Math  /  Calculus

Question(1 point)
Use the Fundamental Theorem of Calculus to find 116sin(x4)x34dx=\int_{1}^{16} \frac{\sin (\sqrt[4]{x})}{\sqrt[4]{x^{3}}} d x= \square

Studdy Solution

STEP 1

What is this asking? We need to calculate a definite integral involving a tricky-looking function using the Fundamental Theorem of Calculus! Watch out! Don't forget to adjust the bounds of integration when making a u-substitution!

STEP 2

1. Substitute
2. Integrate
3. Evaluate

STEP 3

Let's **tackle** this integral using a *u-substitution*!
We see a x4\sqrt[4]{x} term inside the sine function and something similar in the denominator.
Let's set u=x4=x14u = \sqrt[4]{x} = x^{\frac{1}{4}}.

STEP 4

Now, we need to find dudx\frac{du}{dx}.
Remember, the derivative of xnx^n is nxn1nx^{n-1}.
So, dudx=14x141=14x34=14x34\frac{du}{dx} = \frac{1}{4}x^{\frac{1}{4} - 1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}}.

STEP 5

We want to express the integral in terms of uu, so we need to solve for dxdx.
Multiplying both sides of the previous equation by 4x344x^{\frac{3}{4}} gives us 4x34du=dx4x^{\frac{3}{4}}du = dx.

STEP 6

Let's **rewrite** our integral: 116sin(x4)x34dx=116sin(x14)x34dx=116sin(u)x344x34du \int_{1}^{16} \frac{\sin (\sqrt[4]{x})}{\sqrt[4]{x^{3}}} dx = \int_{1}^{16} \frac{\sin (x^{\frac{1}{4}})}{x^{\frac{3}{4}}} dx = \int_{1}^{16} \frac{\sin(u)}{x^{\frac{3}{4}}} \cdot 4x^{\frac{3}{4}} du

STEP 7

Notice how the x34x^{\frac{3}{4}} terms divide to one, which simplifies things nicely!
We're left with: 4sin(u)du \int 4\sin(u) du Don't forget to **change the limits of integration**!
When x=1x = 1, u=14=1u = \sqrt[4]{1} = 1.
When x=16x = 16, u=164=2u = \sqrt[4]{16} = 2.
Our integral becomes: 124sin(u)du \int_{1}^{2} 4\sin(u) du

STEP 8

The integral of sin(u)\sin(u) is cos(u)-\cos(u).
So, we have: 124sin(u)du=412sin(u)du=4[cos(u)]12 \int_{1}^{2} 4\sin(u) du = 4 \int_{1}^{2} \sin(u) du = 4[-\cos(u)]_{1}^{2}

STEP 9

Let's **evaluate** the expression at the upper and lower limits of integration: 4[cos(2)(cos(1))]=4[cos(2)+cos(1)]=4[cos(1)cos(2)] 4[-\cos(2) - (-\cos(1))] = 4[-\cos(2) + \cos(1)] = 4[\cos(1) - \cos(2)]

STEP 10

The value of the definite integral is 4[cos(1)cos(2)]4[\cos(1) - \cos(2)].

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