Calculus
Topics

# How do you use the second fundamental theorem of Calculus to find the derivative of given int sqrt(x^3-2x+6) dx from [-2, x^2]?

Mar 27, 2016

Explained below

#### Explanation:

The second fundamental theorem says this

Now letting $x = t$ the given integral becomes ${\int}_{2}^{{x}^{2}} \sqrt{{t}^{3} - 2 t + 6} \mathrm{dt}$

Accordingly $\frac{d}{\mathrm{dx}} {\int}_{-} {2}^{{x}^{2}} \sqrt{{t}^{3} - 2 t + 6} \mathrm{dt} = 2 x \sqrt{{x}^{6} - 2 {x}^{2} + 6}$ [Put t= ${x}^{2}$ then $\frac{\mathrm{dt}}{\mathrm{dx}}$ = 2x

##### Impact of this question
138 views around the world
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9832984805107117, "perplexity": 2472.616991050206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257605.76/warc/CC-MAIN-20190524104501-20190524130501-00506.warc.gz"}