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Chapter 4: Q. 13 (page 362)

Show that $F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1)$ is an anti-derivative of $f (x) = x^{3} e^{(x^{2})} .$

Short Answer

Expert verified

It is shown that $F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1) is an anti-derivative of f (x) = x^{3} e^{(x^{2})}$ .

Step by step solution

01

Step 1. Given Information.

The anti derivative is $F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1) of f (x) = x^{3} e^{(x^{2})}$ .

02

Step 2. Conclusion.

On differentiating the function, we get,

$\frac{d}{d x} x^{3} e^{(x^{2})} = 2 x^{2} e^{(x^{2})} - e^{(x^{2})} = 2 e^{(x^{2})} (x^{2} - 1) S o, 鈭� x^{3} e^{(x^{2})} d x = \frac{1}{2} e^{(x^{2})} (x^{2} - 1) + C$

Therefore,

$F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1) is an anti-derivative of f (x) = x^{3} e^{(x^{2})} .$

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Most popular questions from this chapter

Prove Theorem 4.13(b): For any real numbers a and b, we have ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ . Use the proof of Theorem 4.13(a) as a guide.

If f is negative on [鈭�3, 2], is the definite integral ${鈭�}_{- 3}^{2} f (x) d x$ positive or negative? What about the definite integral 鈭� ${鈭�}_{- 3}^{2} f (x) d x$ ?

Read the section and make your own summary of the material.

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its value

$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{k^{3}}{n^{4} + n + 1}$

Prove part (b) of theorem 4.4 in the case when n is even: if n is a positive even integer, then ${鈭�}_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

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