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Chapter 2: Problem 9

Find an antiderivative \(\mathrm{F}\) of \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2} \sin \left(\mathrm{x}^{2}\right)\) such that \(\mathrm{F}(1)=0\).

Short Answer

Expert verified
Question: Find the antiderivative \(F(x)\) of the function \(f(x) = x^2\sin(x^2)\), such that \(F(1) = 0\). Answer: \(F(x) = -\frac{1}{2}\cos(x^2) + \frac{1}{2}\cos(1)\).

Step by step solution

01

Integrate the function \(f(x)\)

To find the antiderivative \(F(x)\), we need to integrate the function \(f(x)\). In this case, we will use the technique of substitution. Let \(u = x^2\). Then, \(\frac{du}{dx} = 2x\), which means \(dx = \frac{du}{2x}\). The integral is now: $$\int x^2 \sin(u) \cdot \frac{du}{2x}$$ The \(x\) terms cancel out, leaving us with: $$\frac{1}{2} \int \sin(u) du$$ Now, we integrate the sine function: $$\frac{1}{2} (-\cos(u)) + C$$ We can now substitute back the \(u\): $$F(x) = -\frac{1}{2}\cos(x^2) + C$$
02

Find the value of the constant C

We are given the condition that \(F(1) = 0\). We can now use this information to find the value of the constant C: $$F(1) = 0$$ $$0 = -\frac{1}{2}\cos(1^2) + C$$ Therefore: $$C = \frac{1}{2}\cos(1)$$
03

Write the final antiderivative with the constant C

Now that we have found the value of C, we can write the final antiderivative: $$F(x) = -\frac{1}{2}\cos(x^2) + \frac{1}{2}\cos(1)$$ So, the antiderivative \(F(x)\) is: $$F(x) = -\frac{1}{2}\cos(x^2) + \frac{1}{2}\cos(1)$$

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