Question

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

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

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 \(u\): $$F(x) = -\frac{1}{2}\cos(x^2) + C$$

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: $$C = \frac{1}{2}\cos(1)$$

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: $$F(x) = -\frac{1}{2}\cos(x^2) + \frac{1}{2}\cos(1)$$