Evaluate the definite integral of a piecewise function over two intervals.

Question

NOTE: I need help in solving this integral. I am new to calculus and this expression looks to complicated and I am unable to simplify it.

Solve: $$\int\limits_{x=0}^{ \sqrt[4]{\frac12}} \left( \sqrt[4]{2} x - \sqrt[4]{\frac12} x \right) \, dx + \int\limits_{x= \sqrt[4]{\frac12}}^{1} \left( 1 - \sqrt[4]{\frac12} x \right) \, dx$$

MY ATTEMPT

To solve the given integral expression, we first need to simplify the integrands and then integrate each part separately. Let's start step by step:

1. Simplify the first integrand: $$\sqrt[4]{2}x - \sqrt[4]{\frac{1}{2}}x$$

Factor out the common term $x$: $$x\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right)$$

2. Simplify the second integrand: $$1 - \sqrt[4]{\frac{1}{2}}x$$

Now, we can perform the integration over the specified intervals:

1. For the first integral: $$\int_{x=0}^{\sqrt[4]{\frac{1}{2}}} x\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) dx$$

2. For the second integral: $$\int_{x=\sqrt[4]{\frac{1}{2}}}^{1} \left(1 - \sqrt[4]{\frac{1}{2}}x\right) dx$$

Now, let's compute each integral separately:

1. For the first integral: $$\int_{x=0}^{\sqrt[4]{\frac{1}{2}}} x\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) dx$$

Integrate with respect to $x$: $$\left[\frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) x^2\right]_{x=0}^{x=\sqrt[4]{\frac{1}{2}}}$$

Evaluate the definite integral: $$\frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) \left(\sqrt[4]{\frac{1}{2}}\right)^2 - \frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) \cdot 0^2$$

Simplify further: $$\frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) \left(\sqrt\frac{1}{2}\right)$$

2. For the second integral: $$\int_{x=\sqrt[4]{\frac{1}{2}}}^{1} \left(1 - \sqrt[4]{\frac{1}{2}}x\right) dx$$

Integrate with respect to $x$: $$\left[x - \frac{\sqrt[4]{\frac{1}{2}}}{2}x^2\right]_{x=\sqrt[4]{\frac{1}{2}}}^{x=1}$$

Evaluate the definite integral: $$\left(1 - \frac{\sqrt[4]{\frac{1}{2}}}{2} \cdot 1^2\right) - \left(\sqrt[4]{\frac{1}{2}} - \frac{\sqrt[4]{\frac{1}{2}}}{2} \cdot \left(\sqrt[4]{\frac{1}{2}}\right)^2\right)$$

Simplify further: $$1 - \frac{\sqrt[4]{\frac{1}{2}}}{2} - \left(\sqrt[4]{\frac{1}{2}} - \frac{\sqrt[4]{\frac{1}{2}}}{2} \cdot \sqrt\frac{1}{2}\right)$$

from here I don't how to simplify both expressions. I have numerous times but my answer is wrong everytime. Either I am making mistake in solving integral process or simple arthematic operations.

Answer 1

Since you're asking on how to simplify the expression, here you go:

Call

$$I_1 = \int_{x=0}^{\sqrt[4]{\frac{1}{2}}} x\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) dx$$

and

$$I_2 = \int_{x=\sqrt[4]{\frac{1}{2}}}^{1} \left(1 - \sqrt[4]{\frac{1}{2}}x\right) dx$$

From your steps, we know that

$$ I_1 = \frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) \left(\sqrt\frac{1}{2}\right) $$

$$ I_2 = 1 - \frac{\sqrt[4]{\frac{1}{2}}}{2} - \left(\sqrt[4]{\frac{1}{2}} - \frac{\sqrt[4]{\frac{1}{2}}}{2} \cdot \sqrt\frac{1}{2}\right) $$

We want to find $I_1 + I_2$.

$$I_1 + I_2 = \frac{1}{2}\left(\sqrt[4]{2} - \sqrt[4]{\frac{1}{2}}\right) \left(\sqrt\frac{1}{2}\right) + 1 - \frac{\sqrt[4]{\frac{1}{2}}}{2} - \left(\sqrt[4]{\frac{1}{2}} - \frac{\sqrt[4]{\frac{1}{2}}}{2} \cdot \sqrt\frac{1}{2}\right)$$

Rewrite the above in terms of exponents:

$$I_1 + I_2 = \frac{1}{2} \left(2^{1/4} - 2^{-1/4}\right)(2^{-1/2}) + 1 - \frac{2^{-1/4}}{2} - \left(2^{-1/4} - \frac{2^{-1/4}}{2} \times 2^{-1/2} \right) $$

Simplify the above:

$$I_1 + I_2 = \frac{1}{2} \left(2^{-1/4} - 2^{-3/4} \right) + 1 - \frac{2^{-1/4}}{2} - 2^{-1/4} + \frac{2^{-3/4}}{2}$$

$$I_1 + I_2 = \frac{2^{-1/4}}{2} - \frac{2^{-3/4}}{2} + 1 - \frac{2^{-1/4}}{2} - 2^{-1/4} + \frac{2^{-3/4}}{2}$$

$$I_1 + I_2 = 1 - 2^{-1/4}$$

That gives me the right answer, about $0.1591$.